9

CHAPTER 1 – FUNCTIONS AND RELATIONS IN MATHS YOU CANNOT

• Divide by 0 • Take the square root of a negative number • Take the log of a negative number or 0

SETS TERMINOLOGY eg. Set A = { }7,5,4,3,2,1 Set B = { }10,7,6,2

A∈7 this means that 7 is an element of (or is in) Set A A∉9 this means that 9 is not an element of (or is not in) Set A

{ } A⊆3,2 this means that { }3,2 is a subset of (or little part of) Set A { } A⊆9,2 this means that { }9,2 is not a subset of (or not a little part of) Set A φ this means an empty set (nothing in it) UNION, INTERSECTION AND COMPLEMENT I = Intersection = what is in both (Bucket turned over so small amount) U = Union = what is in either (Bucket upright so large amount) = Set Difference, Elements in the first set but also not in the second. Eg. A B = { }5,4,3,1 B A = { }10,6 COMMON SETS OF NUMBERS Real Numbers (R) All numbers Natural Numbers (N) { }..........4,3,2,1 Integers (Z) { }....2,1,0,1,2..... −−

Rational Numbers (Q) Any number of the form qp which are recurring or terminating

decimals SPECIAL SETS

),( ba The numbers in-between a and b (But not equal to a and b) ],[ ba The numbers equal to and in-between a and b ],( ba The numbers in-between a and b, and also b. On a number line you put a

solid dot at b and a white dot at a ),[ ∞a All numbers larger than a. Note: ∞ always has a ) next to it because you can

never get ∞ exactly. R+ All positive numbers (not including 0) R- All negative numbers (not including 0) R {0} All numbers except for zero

10

RELATIONS AND FUNCTIONS Ordered Pair ),( yx Relation A set of ordered pairs or a rule. Domain : The values of x that the function is valid for. If all values write

Domain = R, if everything but 0, write Domain = R {0} Range : The values of y that the function is valid for. If only positive values

write Range = ),0[ ∞ Function For any function, each x value can only give one y value. If any

vertical line on the graph hits a graph more than once, it is not a function.

1 to 1 Function If any horizontal line on the graph hits a graph more than once, it is not a one to one function.

Image Value of y if you are given x (ie the second number in the bracket) Pre Image Value of x if you are given y (ie the first number in the bracket) Ways of Writing Functions

}:),{( 2xyyx = This has a domain = R, and range [0,∞) or it can also be written 2)(,]5,3[: xxfRf =→− . The only difference is the second one is only graphed or

considered when x is between -3 and 5 inclusive. Implied/Maximal Domain When the domain is not given, the implied/maximal domain is the largest set of numbers

which the function is valid for. For example for x

xf 1)( = , the function works for all values

of x except when x=0. Therefore the maximal domain is R {0} Odd Functions Have the property that )()( xfxf −=− . Even Functions Have the property that )()( xfxf =− . This means on either side of the y-axis they are symmetric. Modulus Function x is the absolute value of a function, making the result always positive. ie 34− = 34

When sketching the graphs make a hybrid function (one rule for certain values of x, and another function for the rest). Eg. 23)( −+= xxf The modulus is not needed if 3−≥x as (x+3) is

already positive. So 23)( −+= xxf if 3−≥x When 3−≤x , the modulus function is multiplying (x+3) by -1 to make it positive.

So 23)( −−−= xxf if 3−≤x Sums of Functions

)()())(( xgxfxgf +=+ but the function (f+g)(x) is only valid where both f(x) and g(x) were both valid together. ie Intersection of domains.

11

Products of Functions )()())(( xgxfxfg = but the function (fg)(x) is only valid where both f(x) and g(x)

were both valid together. ie Intersection of domains. Composite Functions

)]([)( xgfxgf =o This looks like Therefore if the Range of g is a subset of the domain of f, the function fog(x) is defined. ie )(xgf o needs ran g ⊆ dom f. Eg. xxf =)( and 82)( −= xxg Make a table Since ran g ⊆ dom f. therefore )(xgf o does not exist. The only way to make it

exist is to change the domain of g to above 4 making only positive numbers in the range.

Note : Much of this is common sense, 82)]([ −= xxgf which will not work if x is less

than 4 because you cannot take the square root of a negative number. The mathematical way to say this is ran g ⊆ dom f.

Inverse Functions These are functions reflected about the line y=x (Points become reversed 3,2 2,3)

• Inverse Functions are only valid for one to one functions (horizontal line test) so you may have to enforce restrictions or make it a hybrid function.

• The dom f = range 1−f and vice versa • To find the inverse equation swap x and y and then rearrange

There are 2 ways to determine the inverse function Method 1 : Swap x and y, then rearrange

Eg 35 −= xy So 35 −= yx Rearrange )3(51

+= xy

Method 2 : Solve xxff =− )]([ 1

Eg xxf =−− 3)(5 1 So )3(51)(1 +=− xxf

Domain Range

g R R

f [0,∞) [0,∞)

g(x) Domain of g

f(x) Range of g Range of f

12

CHAPTER 3 – FAMILIES OF FUNCTIONS DOMAIN AND RANGE Maximal Domain : The values of x that the function is valid for. If all values write Maximal domain = ℜ , if everything but 0, write Maximal Domain = 0ℜ Range : The values of y that the function is valid for. If only positive values write Range =

),0[ ∞ The Range and Domain are affected by the fact that you cannot

• Divide by 0 • Take the square root of a negative number • Take the log of a negative number or 0

TRANSFORMATIONS OF khxay n ++= )( All polynomials in the form khxay n ++= )( have been reflected (flipped), dilated (squashed or stretched) or translated (moved up/down across) from the original graph of

nxy = , where n can be any number. See the Common Graphs page (page 4) to see all the original graphs in the Maths Methods Course. To determine the transformation equation (By ALGEBRA) Determine the how a point ),( yx becomes the new point ),( yx ′′ Note: The following transformations can occur to the x values.

• Reflection in the y-axis [or )( xfy −= ] xx −→′

• Dilation of factor m from the y-axis [or )(mxfy = ] mxx →′

• Translate n units in the positive direction of the x-axis or nxx +→′ Translate n units parallel to the x-axis [or )( nxfy −= ]

Note: The following transformations can occur to the y values.

• Reflection in the x-axis [or )(xfy −= ] yy −→′ • Dilation of factor m from the x-axis [or )(xmfy = ] myy →′ • Translate n units in the positive direction of the y-axis or nyy +→′

Translate n units parallel to the y-axis [or nxfy += )( ] STEPS

Once you have completed these transformations in the correct order you will get 2 equations )(xgx =′ and )(yhy =′ .

Rearrange each into =x and =y substitute these into the original nxy = to get the new transformation equation. Finally change x′ to x and y′ to y

New x values are the oldones multiplied by -1

13

Eg Find the rule of the image when the graph of 2

1x

y = is translated 4 units in the

negative direction of the x-axis, reflected in the y-axis and dilated a factor of 3 from the x-axis

ANS Translate 4 units in the negative direction of the x-axis 4−→′ xx Reflected in the y-axis )4( −−→′ xx Dilated a factor of 3 from the x-axis yy 3→′ So the 2 equations are 4+−→′ xx and yy 3→′

Rearrange these to 4+′−→ xx and yy ′→31

Original is 2

1x

y =

2)4(1

31

+′−=′

xy so 2)4(

3+′−

=′x

y and changing back 2)4(3+−

=x

y

ADDITION OF ORDINATES To sketch the graph of ))(( xgf + just sketch )(xf and then )(xg . Then add the heights of each graph at any x point to sketch the new graph.

14

GRAPHICAL TRANSFORMATIONS (Use this table if instructions given in the correct order Sideways movement must be in correct order = x coordinates = in bracket = Translate, dilate, reflect. Up/down movement must be in correct order = y coordinates = out of bracket = Reflect, dilate, translate

y = -** k f ( -* ax - b ) + m

ORDER EQUATION OFFICIAL TRANSFORMATION

DRAWING (dotted is original)

UNOFFICIAL EXPLANATION

EFFECT ON OLD CO-ORDINATE

Sideways Movement (Translate, dilate, reflect)

First b

(in bracket)

Translate “b” units in the positive direction of the x-

axis or Translate “b” units parallel to x-axis

Moves it right. Add b to each x–

coordinate (x + b, y)

Second a (in front of x)

Dilated a factor of a1 from

the y-axis

If a > 1 thinner If a < 1 wider

(height is same)

Each x-coordinate multiplied by 1/a

(x/a, y)

Third -* (one in bracket) Reflection in the y-axis

Turns it side on Each x-coordinate multiplied by –1

(-x, y) Up/down Movement (Reflect, dilate, translate)

First -** (one at front) Reflection in the x-axis

Turns it upside down

Each y-coordinate multiplied by –1

(x, -y)

Second k (at the front)

Dilated a factor of k from the x-axis

If k > 1 taller If k < 1 shorter

(width the same)

Each y-coordinate multiplied by k

(x, ky)

Third m (at the back)

Translated “m” units in the positive direction of the y-

axis or Translate “m” units parallel to y-axis

Moves it up Add m to each y-

coordinate (x, y + m)

If told instructions in different order, use the algebra method on the previous page or override previous instructions (see last Page)

15

COMMON GRAPHS

y = sin(x)Important aspects

Amp = 1, Period = 2π Starts at origin (0,0)

y = cos(x)Important aspects

Amp = 1, Period = 2π Starts at origin (0,0)

y = tan(x)Important aspects

Asymptotes at x = 2π

− , 2π

Goes thru origin, Period = π

TRANSFORMATIONS

Eg. y = - 3 (- 2x + 1)2 + 4

SIDE TO SIDE MOVEMENT (Brackets) • Move left 1 (+1 in bracket) translation of -1 units parallel to the x-axis

or also written translation of 1 units in negative direction of x-axis • dilate by factor ½ from y-axis (squish in) – taken from the 2 • reflect about y-axis (negative in bracket)

UP AND DOWN MOVEMENT (Outside Brackets)

• reflect about x-axis (negative at front) • dilate by factor 3 from x-axis (stretch out) – taken from the 3 • Move up 4 (+4 at end) translation of 4 units parallel to the y-axis or

also written translate 4 units in positive direction of y-axis

xy 1=

Important aspects Asymptotes x = 0 and y = 0 Dom and Range = R / {0}

2

1x

y =

Important aspects Asymptotes x = 0 and y = 0 Dom = R / {0}, range = R+

y = ex Important aspects Asymptote of y = 0

Goes through (0,1) and Range = (0,∞)

y = loge(x)Important aspects Asymptote of x = 0 Goes through (1,0)

Domain = (0,∞)

y = x

Important aspects Turning Point (0,0)

Range = [0,∞)

y = ax + bImportant aspects

Gradient = a, y-int = (0,b)

y = x2

Important aspects Turning Point (0,0)

Range = [0,∞)

y = x3

Important aspects Point of inflect : (0,0)

y = x4

Important aspects Turning Point (0,0)

Range = [0,∞)

xy =

Important aspects Starting Point (0,0)

Domain and Range = [0,∞)

5

EXAMPLES EXAMPLE ONE – In Correct Order Find the rule of the image when the graph of xy = is in order

translated 4 units in the negative direction of the x-axis, reflected in the x-axis, dilated by factor 2 from the x-axis, dilated by factor ½ from the y-axis, reflected in the y-axis, translated 3 units in the positive direction of the y-axis

ANS First check the sideways (bracket) movement, is it in the correct order (translate, dilate, reflect)

translated 4 units in the negative direction of the x-axis means + 4 in bracket

dilated by factor ½ from the y-axis means 2 in front of x

reflected in the y-axis means negative in bracket

YES it is, So far the graph is )42( +−= xy Now check the up/down (outside bracket) movement, is it in the correct order (reflect, dilate, translate)

reflected in the x-axis means negative at the front

dilated by factor 2 from the x-axis means a 2 at the front

translated 3 units in the positive direction of the y-axis means a 3 at the end

YES it is, So the new rule is 3)42(2 ++−−= xy

6

EXAMPLE TWO – Where not given in correct order

Find the rule of the image when the graph of x

y 1= is in order

translated 1 unit in the negative direction of the y-axis dilated by factor ¼ from the y-axis dilated by factor 3 from the x-axis translated 2 units in the positive direction of the x-axis

First check the sideways (bracket) movement, is it in the correct order (translate, dilate, reflect)

dilated by factor ¼ from the y-axis translated 2 units in the positive direction of the x-axis

NO, so we have to do ALGEBRA

dilated by factor ¼ from the y-axis xx41

→′

translated 2 units in the positive direction of the x-axis 241

+→′ xx

translated 1 unit in the negative direction of the y-axis 1−→′ yy dilated by factor 3 from the x-axis 33)1(3 −→−→′ yyy

So the 2 equations are 241

+→′ xx and 33 −→′ yy

Rearrange these to 84 −′→ xx and 131

+′→ yy

Original is x

y 1=

)84(

1131

−′=+′

xy

1)84(

131

−−′

=′x

y

so 3)84(

3−

−′=′

xy and changing back 3

)84(3

−−

=x

y

OR A SHORTCUT

translated 1 unit in the negative direction of the y-axis means -1 at end dilated by factor ¼ from the y-axis means 4 in front of

x dilated by factor 3 from the x-axis means 3 at the front translated 2 units in the positive direction of the x-axis means -2 in bracket

Since this is not in the correct order we use brackets

translated 1 unit in the negative direction of the y-axis 11−=

xy

dilated by factor 3 from the x-axis )11(3 −=x

y

33−=

xy

{Note we put brackets around everything because we are doing things in wrong order}

7

dilated by factor ¼ from the y-axis x4 in bracket translated 2 units in the positive direction of the x-axis )2(4 −x 84 −x

in bracket

Combining 3)84(

3−

−=

xy

CHAPTER 4 - POLYNOMIALS POLYNOMIALS (Ex 4A and A3 page 599) The highest power (after being expanded) is known as the degree of the polynomial. Eg 24)( += xxf Degree One (Linear) 423)( 2 ++= xxxf Degree Two (Quadratic) 237)( 23 +−−= xxxxf Degree Three (Cubic)

72)( 34 −+= xxxf Degree Four (Quartic) EXPANDING BRACKETS

The binomial theorem is used to expand (a+b)n The value of each term is given by n nCr r rrn ba − All you need to determine is the value of a, b, r (the power of b) and n The tricky part is that the 1st term is r=0, 2nd term is r = 1 etc.

Note : nCr button is in MATH PRB : 3 or can be seen from Pascals triangle

Eg Find the coeffeient of x7 in the expansion of (2x – 5)10 Here n = 10, a = 2x, b = -5 For x7 this will be the term (2x)7(-5)3 so r = 3 Using the formula n nCr r rrn ba − = 10 nCr 3 (2x)7(-5)3 = 120 (2x)7(-5)3 = 120 (128) (-125) x7 = -1 920 000 x7 so the coefficient is -1 920 000 Eg Find the 5th term of the expansion of (3x+4)7

5 nCr 3 = 10 6th row, 4th number

4th row, 2nd number 3 nCr 1 = 3

8

Here n = 7, a = 3x, b = 4, r = 4 (the 5th term – note the tricky part) Using the formula n nCr r rrn ba − = 7 nCr 4 (3x)3(4)4 = 35 (3x)3(4)4 = 35 (27) (256) x3 = 241 920 x3

9

EQUATIING COEFFICIENTS Expand the brackets and match the coefficients of 3210 ,,, xxxx etc Eg If bxaxx ++=−+ 22 )3(7122 , determine the values of a and b

Expand the right hand side )9(6)96( 22 baaxaxbxxa +++=+++

Now match the coefficients 2x 222 axx = 1x axx 612 = 0x ba +=− 97

Therefore a=2 , a612 = , 79 −=+ ba so 2=a and 7)2(9 −=+ b , 25−=b

DIVISION OF POLYNOMIALS The following example shows how to divide polynomials. In the example below 2−x is the divisor and 33 3 −+ xx is known as the dividend. Eg Divide 33 3 −+ xx by 2−x The Box Method So the final answer is 1363 2 ++ xx with a remainder of 23

Note : 2

231363 2

−+++

xxx

3 0 1 -3

-

3 0 1 -3

- 3 x 2 = 6 6 x 2 = 12 13 x 2 = 26

3 + 0 = 3 0 + 6 = 6 1 + 12 = 13

-3 + 26 = 23

2 In the box goes the solution to the divisor equals 0 ie x-2 = 0 x = 2

Coefficients of the dividend

3103 23 −++ xxx

This will be the remainder

2

Times the previous bottom row by box

Always a dash goes under here

Bottom row is add rows 1 and 2

10

REMAINDER THEOREM A big shortcut is that when dividing a polynomial )(xP the remainder is obtained by substituting the solution of the divisor = 0 Eg Divide 33 3 −+ xx by 2−x divisor = 0 [ x-2 = 0, x = 2 ]

So the remainder is given by 2332243)2()2(3)2( 3 =−+=−+=P FACTOR THEOREM

bax + is a factor of )(xP if and only if the remainder is 0 when the solution to ax+b=0 is substituted into the polynomial. Example Is x – 3 a factor of 3102 3 −− xx The solution of the divisor x = 3, then substitute into 3102 3 −− xx

213)3(10)3(2 3 =−− Since the remainder is not equal to 0, x – 3 is not a factor FACTORISING POLYNOMIALS Use calculator (table of values) to find where 0)( =xP , this gives factors and you can use division to find the others. Eg. Factorise 108912193 +−− xxx

First put this function into calculator and look at the table of values, you will find that the polynomial = 0 when x = -11, 9 and 11. Therefore the 3 factors are )11)(9)(11( −−+ xxx

X-AXIS INTECEPT(S) Value(s) of x when y=0 Y-AXIS INTERCEPT Value of y when x = 0

11

QUADRATIC FUNCTIONS (Ex 4B) USE OF CALCULATOR • Press Y = , then enter the equation • Then press GRAPH (adjust WINDOW if you cant see the graph) • Y – INTERCEPT

Press TABLE (2nd GRAPH) and read the value of Y1 when X = 0 • X – INTERCEPTS

Press GRAPH, then CALC, then ZERO and find both • NATURE OF TURNING POINT

- MAXIMUM if graph has angry face this is where the number in front of x2 is negative (unhappy) - MINIMUM if graph has happy face this is where the number in front of x2 is positive (happy)

• TURNING POINT Press GRAPH, then CALC, then Maximum or Minimum

• AXIS OF SYMMETRY x = first number of turning point

Example For the equation 452 ++= xxy Use the table and graph to see The y-intercept is (0,4)

The x-intercepts are (-4,0) and (-1,0) The turning point is a Minimum The turning point is (-2.5,-2.25) The axis of symmetry is x = -2.5 The mimimum value of is -2.25 (the lowest the graph goes)

1. y - intercept

3. Turning Point (sometimes also known as vertex)

2. x - intercepts

4. Axis of symmetry

12

SHORTCUTS If the equation is of the form cbxaxy ++= 2 a) The y-intercept is equal to (0,c) b) The x-intercepts are calculated using the quadratic formula

a

acbbx2

42 −±−=

acb 42 − is known as the determinant, if the determinant is positive there are 2 x-intercepts, if it is equal to 0 there is only 1 x-intercept (or solution) and if the determinant is negative there are no x-intercepts (or solutions to the equation)

c) The axis of symmetry is abx

2−

=

d) The turning point is equal to ,2

(ab− substitute

abx

2−

= into the equation)

EXAMPLE

452 ++= xxy this is of the form cbxaxy ++= 2

So the y-intercept is (0,4) and the axis of symmetry 5.2)1(2

52

−=−

=−

=abx

The turning point will be a minimum (number in front of x2 is positive = smiling) The value of the turning point is (-2.5, (-2.5)2+5(-2.5)+4 ) = (-2.5, -2.25)

The x-intercepts are 2

95)1(2

)4)(1(455 2 ±−=

−±−=x

Which are x = (-4,0) and (-1,0) If the equation is of the form khxay ++= 2)( a) To get the y-intercept substitute x = 0 b) To get the x-intercepts, best to expand into the form cbxaxy ++= 2 , then use the

quadratic formula c) The turning point is given by ),( kh− Example

5)3(2 2 +−−= xy this is of the form khxay ++= 2)( So the y-intercept is (0, -2(0-3)2 + 5) = (0,-13) The turning point is (3,5) (opposite of number in bracket, number at the end) If the equation is of the form ))(( exdxay ++= a) To get the y-intercept substitute x = 0 b) The x-intercepts are )0,( d− and )0,( e− c) To get the turning point, best to expand into the form cbxaxy ++= 2 , then use

abx

2−

=

Example

)2)(3(2 +−= xxy this is of the form ))(( exdxay ++= So the y-intercept is (0, 2(0-3)(0+2)) = (0,-12) The x-intercepts are (3,0) and (-2,0) (opposite of number in brackets)

13

Converting Quadratics from cbxaxy ++= 2 into khxay ++= 2)( This can be shown by the following example. Eg 45123 2 +−= xxy )154(3 2 +− xx Determine the value of new x coefficient divided by 2 (in this case = -2)

]15)2()2[(3 22 +−−−x [ ])11)2(3 2 +−x

33)2(3 2 +−x Converting Quadratics from cbxaxy ++= 2 into ))(( exdxay ++= This can be shown by the following example. Eg 11132 2 +−= xxy

2

)22)(112( −− xx

2)1(2)112( −− xx

)2)(112( −− xx DETERMINING THE EQUATION OF A PARABOLA (Ex 4C) 1. If you are given the turning point (A)

Eg Turning point = (6,-3) Put into turning point form 3)6( 2 −−= xay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

2. If you are given the 2 x-intercepts (B)

Eg x-intercepts (-3,0) and (5,0) Put into x-intercept form )5)(3( −+= xxay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

3. If you are given the y-intercept and 2 other points

Eg y-intercept (0,6) Put into y-intercept form cbxaxy ++= 2 and c = 6 in this case

Then substitute x and y from the other 2 point on the graph. This gives you 2 equations, 2 unknowns and you can solve for a and b.

Factorise a from the rest of equation

Put this number in bracket with x, subtract this number squared outside.

Opposite of first number of turning point

Second number of turning point

Above the last term multiply a by c Now find the factors that add to give -13 and multiply to 22 (-11 and -2)

22

x– 1 0 – 5 5 1 0

y

– 1 0

– 5

5

1 0

x– 1 0 – 5 5 1 0

y

– 2 0

– 1 0

1 0

2 0 A) B)

Multiply big [ ] brackets out

14

CUBIC FUNCTIONS (Ex 4E) If the coefficient of 3x is positive then cubic graphs are below the x-axis on the left hand side of the graph and above the y-axis on the right hand side of the graph. There are 3 basic types of cubic graphs a) khxay ++= 3)( This has a point of zero gradient at ),( kh− and only 1

x-intercept. b) )()( 2 fxexay ++= This has 2 turning points and 2 x-intercepts that are (-

e,0) and (-f,0) (Note : (-e,0) is also one of the turning points) c) ))()(( jxhxgxay +++= This has 2 turning points and 3 x-intercepts that are (-

g,0), (-h,0) and (j,0). SIGN DIAGRAMS Number lines which show if an expression is positive or negative. 3 examples are shown above. Use calculator or factorise to get the x-intercepts so a Sign Diagram can be drawn. DETERMINING THE EQUATION OF A CUBIC (Ex 4G) 1. If you are given the point of zero gradient and one other point. (A)

Eg Point of zero gradient = (3,-4) Put into zero gradient form 4)3( 3 −−= xay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

2. If there are only 2 X-Intercepts and one is also a turning point (B)

Eg x-intercepts (-5,0), (4,0) and (-5,0) is also a turning point. Put into form )4()5( 2 −+= xxay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

3. If you are given 3 X-intercepts and one other point (C)

Eg x-intercepts (-3,0), (-1,0) and (5,0) Put into x-intercept form )5)(1)(3( −++= xxxay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

4. If you are given the y-intercept and 3 other points

Eg y-intercept (0,6) Put into y-intercept form dcxbxaxy +++= 23 and d = 6 in this case

Then substitute x and y from the other 3 points on the graph. This gives you 3 equations, 3 unknowns and you can solve for a, b and c.

x– 1 0 – 5 5 1 0

y

– 1 0

– 5

5

1 0

x– 1 0 – 5 5 1 0

y

– 2 0

– 1 0

1 0

2 0

x– 1 0 – 5 5 1 0

y

– 1 0

– 5

5

1 0 A) B) C)

+ + +

15

QUARTIC FUNCTIONS (Ex 4F and 4G) Quartic functions )( 4x have a similar shape to parabolas but can have up to 3 turning points. They can have 0,1,2,3 or 4 x-intercepts as shown in the examples below. Note: Even powered polynomias )( 2x and )( 4x can have no x-intercepts but all odd powered polynomials must hit the x-axis at least once. DETERMINING THE EQUATION OF A QUARTIC 1. If you are given the turning point and one other point. (A)

Eg Turning Point = (2,3) Put into turning point form 3)2( 4 +−= xay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

2 If you are given 2 x-intercepts which are also turning points (B)

Eg x-intercepts (-3,0), (2,0) which are also turning points Put into form 22 )2()3( −+= xxay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

3. If you are given 3 x-intercepts and one is also a turning point (C)

Eg x-intercepts (-1,0), (-2,0) and (3,0) which is also a turning point. Put into form 2)3)(2)(1( −++= xxxay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

4. If you are given 4 x-intercepts (D)

Eg x-intercepts (-1,0), (-2,0), (3,0) and (4,0) Put into form )4)(3)(2)(1( −−++= xxxxay

Then substitute x and y from any other point on the graph into the equation to find the value of a.

5. If you are given the y-intercept and 4 other points

Eg y-intercept (0,6) Put into y-intercept form edxcxbxaxy ++++= 234

and e = 6 in this case Then substitute x and y from the other 4 points on the graph. This gives you 4 equations, 4 unknowns and you can solve for a, b, c and d.

BIG EXAM TRICK If the equation is )4)(3)(2)(1( −−++−= xxxxy sometimes the negative at the front is removed by multiplying one of the brackets by -1 eg. )3()3( xx −=−− , so therefore )4)(3)(2)(1( −−++−= xxxxy

can also be written )4)(3)(2)(1( −−++= xxxxy

x– 10 – 5 5 10

y

– 10

– 5

5

10 A) B) C) D)

x– 10 – 5 5 10

y

– 10

– 5

5

10

x– 10 – 5 5 10

y

– 10

– 5

5

10

x– 10 – 5 5 10

y

– 10

– 5

5

10

16

CHAPTER 5 – EXPONENTIALS AND LOGS EXPONENTIAL GRAPHS The general graph of xay = is shown below. The significant points of the graph are the horizontal asymptote at y = 0 and the points (0,1) and (1,a). Since the graph below has a the point (1,2) then a=2. The normal domain = R and the normal range = R+ POWER RULES You may not have to remember all of these as common sense can be used. eg. a2 x a3 = a x a x a x a x a = a5 RULES

1. an x am = an+m 5. n

nn

ba

ba

=⎟⎠⎞

⎜⎝⎛

2. an ÷ am = an-m 6. a0 = 1 3. (an)m = anm 7. aa =5.0 4. (ab)n = anbn 8. a-n = na

1

x– 4 – 2 2 4

y

5

10

15

20

25

30

17

SOLVING FOR UNKNOWN (x) a) UNKNOWN ON BASE (at bottom)

Put both sides to the power of something so you get x1 =, and break big numbers down to prime factors.

Note : When simplifying break numbers down to their lowest power, use the

following. Remember 1 = any number to the power of 0.

4 = 22 36 = 62 144 = 24 . 32 1024 = 210 8 = 23 49 = 72 243 = 35 2048 = 211 9 = 32 64 = 26 256 = 28 2187 = 37 16 = 24 81 = 34 343 = 73 3125 = 55

25 = 52 100 = 102 512 = 29 3401 = 74 27 = 33 125 = 53 625 = 54 4096 = 212 32 = 25 128 = 27 729 = 36

eg x-4 = 16 becomes x-4 = 24 then put both sides to the power of 41

−

gives x1 = (24)-0.25 = 2-1 = 0.5

eg 2723

=x becomes 323

3=x then put both sides to the power of 32

gives x1 = 93)3( 232

3 == b) UNKNOWN ON INDEX (at top)

Expand totally and get bases the same on both sides (If bases can’t be the same you have to take logs or maybe use substitution – see below) eg. 3x = 27 3x = 33 x = 3

eg. 125

15 2 =+x , so you make it 32 55 −+ =x so x = -5.

Some need to be solved by substitution. eg. 027)3(1232 =+− xx Not everything can be put into powers of 3.

Let xy 3= The equation becomes 027122 =+− yy Factorising 0)3)(9( =−− yy so y=9 and y=3 So 93 == xy meaning x=2 and 33 == xy meaning x=1 Using LOGS

Taking Logs of both sides brings anything at the top to the bottom (rule 1 of logs), once the unknown is at the bottom normal maths applies

Eg

2log10log

10log2log10log2log

102

e

e

ee

ex

e

x

x

x

=

==

=

18

LOGS • ax = b x = loga b • Just remember the example 42 = 16 log416 = 2 • Log = what do I put to the power of to get • On Calculator log = log10 and ln = loge

2 Rules of Logs

• logmnx = x logmn (at top can be put to front and vice versa)

• logma + logmb - logmc + logmd = )c

abd(logm

(all adding bits are put on top of fraction, subtract bits on bottom) Changing real numbers to Logs

• 0 = log1 (can be any base) • 1 = log22 = log33 = log44 = log55 = log1010= logee = logaa = logbb • 2 = log24 = log39 = log416 = log525 = log10100= logee2 = logaa2 = logbb2 • 3 = log28 = log327 = log464 = log5125 = log101000= logee3 = logaa3 • 4 = log216 = log381 = log4256 = log5525 = log1010000= logee4 = logaa4 • x = logeex = log1010x

TO SIMPLIFY OR SOLVE

• Must change real numbers to logs so you get log( ) + log( ) - log( ) + log( ) = log ( ) etc (no numbers unless in log brackets)

• Now use the 2 log rules to get log (…) = log(…) then make brackets equal. Eg. log2(2x+1) + 2 log23 – 0.5log225 = 3

Now you must make sure there are no numbers or symbols unless in log brackets log2(2x+1) + log232 – log2250.5 = log28 log2(2x+1) + log29 – log25 = log28 Use log laws to simplify

)8(log)5

)12(9(log 22 =+x

Therefore what’s inside the brackets must be equal

85

918=

+x So 1831

=x

USING CALCULATOR (Change of Base) The calculator only has log = log10 and ln = loge So if you wanted to work out log25, you have to

Use the rule that ax

ax

xe

ea 10log

logloglog

log 10==

So to work out log25, on your calculator work out 32192.22log5log

2log5log

10

10 ==e

e

19

LOGARITHMIC GRAPHS The general graph of xy alog= is shown below. The significant points of the graph are the vertical asymptote at x = 0 and the points (1,0) and (a,1). Since the graph below has a the point (10,1) then a=10. The normal domain = R+ and the normal range = R. It is the inverse function of xay =

DETERMING THE RULE GIVEN A EXPONENTIAL OR LOGARITHMIC GRAPH If there are 2 unknowns (usually a and b), you need 2 points to solve for these. Three examples are shown. 1. Find the equation in the form baey x += if it goes through the points (0,5) and

(2,21).

ANS The equations are bae

bae+=

+=2

0

215 subtracting removes the b’s

)1(16 2 −= ea

8383.01

162 =−

=e

a

From first equation b = 5 – a = 4.1617 2. Find the equation in the form )(log bxay e += if it goes through the points (8,0) and

(12,2). ANS Being told the value of x when y=0 is extremely useful since the only way y

could equal 0 is if it is 0)(log =+ bxe which means (8+b)=1, so b=-7 Then substitute this into the other equation

5log

2)712(log2

e

e

a

a

=

−=

3. Find the equation in the form bxAey = if it goes through the points (1,4) and (3,10).

ANS Write the 2 equations b

b

AeAe

3104

=

=

Divide the bottom by the top to eliminate A

25.2log5.2log2

5.2

2

2

e

b

b

be

=

==

Then substitute into one of the equations to find A

x2 4 6 8 10

y

– 2

– 1.5

– 1

– 0.5

0.5

1

20

CHAPTER 6 – CIRCULAR FUNCTIONS

DEGREES AND RADIANS Use the following chart to convert radians to degrees and vice versa. To change from degrees on calculator put in radians mode (mode, radian), then type 30° on home screen (° is found by 2nd,apps,and press enter on the ° option) THE UNIT CIRCLE (Very Useful to understand and use this) An example of radians, degrees, sine and cosine is below. This is a unit circle with a radius of 1, therefore the perimeter of the circle = 2π(1) = 2π Therefore one full revolution is 360° (degrees) or the end of the arrow travels 2π (radians) Degrees = how many degrees anticlockwise it has moved from the horizontal. Radians = the distance the arrow has moved. Sine = the y-coordinate of where the arrow is now.

(how far it has moved vertically from origin) Cosine = the x-coordinate of where the arrow is now.

(how far it has moved horizontal from origin) For example, the arrow has moved a total of 120°, this means the arrow has travelled a

total of 3

22360120 ππ =x radians. The co-ordinates of the arrow are )

23,

21(− , this means

that sin(120°) = 23 and cos(120°) =

21

−

Degrees

X 180π

Radians

X π

180

x– 1 1

y

– 1

1

0°

90° or 2π

180° or π

270° or 2

3π

120°

SIN

COS

21

SUMMARY OF RESULTS (Must be learnt)

Degree 0° 30° 45° 60° 90° 180° 270° 360°

Radians 0 6π

4π

3π

2π π

23π π2

Sine 0 21

21

23 1 0 1− 0

Cosine 1 23

21

21 0 1− 0 1

Tangent

cossin 0

31 1 3 undef 0 undef 0

Note: Sin / Cos of multiples of 90° or 2π are either 0 or 1±

Sin / Cos of multiples of 45° or 4π are either

21

±

Sin / Cos of multiples of 30° or 6π ,

3π are either

21

± or 23

±

Note : 866.023=

Tan is just Sin divided by Cos Knowing this helps and using the unit circle helps you to determine others (see example below)

Eg )3

5sin( π

We can work out that this is 300°, draw this on the circle Sin/cos multiples of 30° give a

Result of either 21

± or 23

±

And seeing the unit circle will show us which one

From the graph it is obvious

)3

5sin( π where it hits vertical

line so

23)

35sin( −=π

x– 1 1

y

– 1

1

0°

90°

180°

270°

300

SIN

COS

22

SYMMETRIC PROPERTIES Use the unit circle to see these properties. SINE )sin()2sin( θπθ =+ COSINE )cos()2cos( θπθ =+ )sin()sin( θπθ −=+ )cos()cos( θπθ −=+

)cos()2

sin( θπθ =+ )sin()2

cos( θπθ −=+

)sin()2sin( θθπ −=− )cos()2cos( θθπ =− )sin()sin( θθπ =− )cos()cos( θθπ −=−

)cos()2

sin( θθπ=− )sin()

2cos( θθπ

=−

)sin()sin( θθ −=− )cos()cos( θθ =− TAN )tan()2tan( θπθ =+ )tan()tan( θπθ =+

)tan()2tan( θθπ −=− )tan()tan( θθπ −=−

)tan()tan( θθ −=−

Eg. Using unit circle to determine the symmetric property of )2

cos( πθ +

Eg Try θ = 30° The value of

21)

2cos( −=+

πθ

How does this compare to sin

and cos of θ.

Since 21)sin( =θ

The rule must be

)sin()2

cos( θπθ −=+

QUADRANTS

1st QUADRANT 0 to 90° or 0 to

2π

All are positive (sin,cos and tan)

2nd QUADRANT 90 to 180° or

2π to π

Only Sine is positive (vertical axis)

3rd QUADRANT 180 to 270° or π to

23π

Only Tan is positve

4th QUADRANT 270 to 360° or

23π to π2

Only Cos is positive (horizontal axis)

x– 1 1

y

– 1

1

0°

90°

180°

270°

30°

SIN

COS30°

23

SOLVING EQUATIONS WITH SIN / COS / TAN

Use the table / unit circle to determine the value of the brackets Note : Put down the 2 solutions between 0 and 2π, then also the ones in the next

cycles (add 2π, 4π etc) Solve (being careful about the domain)

EXAMPLES

Solve 2

3)12

2( −=−

πθCos

Find the values between 0 and 2π where cos = 2

3−

The dot shows where cos = 2

3−

Which is 6

5π and 6

7π

Next ones would be on the next revolution (+2π)

6

176

126

5 πππ=+ ,

619

612

67 πππ

=+

Next ones are 6

29π and 6

31π etc

So 2

3)12

2( −=−

πθCos

=−12

2 πθ6

5π , 6

7π , 6

17π , 6

19π , 6

29π , 6

31π add 12π to both sides

=θ212

11π , 12

15π,

1235π ,

1239π ,

1259π ,

1263π divide both sides by 2

=θ24

11π , 24

15π , 24

35π , 24

39π , 24

59π , 24

63π

Note : We do not count the last 2 since they are not between 0 and 2π

x– 1 1

y

– 1

1

0°

90°

180°

270°

SIN

COS

24

GRAPHS OF SIN / COS / TAN SIN GRAPH (starts at origin) COS GRAPH (starts at x=0,y=1) Important aspects Important aspects Amp = 1, Period = 2π Amp = 1, Period = 2π Starts at origin (0,0) Starts at origin (0,1) TAN GRAPH Important aspects Asymptotes at x =

2π

− , 2π

Goes thru origin, Period = π

x

y

– 1

– 0.5

0.5

1

x

y

– 1

– 0.5

0.5

1

π− 2π−

2π π

23π π2 π−

2π−

2π π

23π π2

x

y

– 4

– 2

2

4

π− 2π−

2π π

23π π2

25

TRANSFORMATIONS Just think about how these normal graphs are transformed to determine new periods (how long to finish one cycle), amplitudes (how high it goes from normal position) etc or use the rules below SIN, COS For the graphs dcbxay +−= )sin( or dcbxay +−= )cos(

PERIOD = How long it takes to finish one cycle = bπ2 (ie 2π / number next to x)

AMPLITUDE = how high it goes from normal (or middle) position = a The number at the front (Always positive)

RANGE = [ d - a , d + a ] TAN y = a tan (bx-c) + d PERIOD = How long it takes to finish one cycle =

bπ (ie π / number next to x)

DOMAIN = Everywhere but where the vertical asymptotes are b

ckx

+= 2

π

where k is

any odd integer RANGE = R Examples Find the amplitude and period of each of the following a) y = 8 sin (4x + 1) - 3

Think of transformations the +1, moves it left 1 The 4 squashes it in 4 times, meaning the period which is normally 2π for a sine

graph becomes 24

2 ππ=

The 8 stretches it 8 times upwards meaning the amplitude which is normally 1 for a sine graph becomes 8 and the -3 moves the graph up 3, making the range [-5,11]

b) y = 4 tan (2x + π)

Think of transformations the +π, moves it left π The 2 squashes it in 2 times, meaning the period becomes which is normally π for

a tan graph becomes 2π ,

The asymptotes are usually 2π

±=x , moved left π moves them to 2π

− , 2

3π− then

the squashing in makes them 4π

−=x , 4

3π−

26

GETTING THE EQUATION OF SIN, COS GRAPHS y = ______ ______ _________ ( ______ x ) + ________ 1. Look for the middle line, this number goes at position 1. 2. SIN if starts (x=0) on middle line, COS if it doesn’t start on middle line, sin or cos

goes in position 2. 3. How high the graph goes from the middle line (the amplitude), this number goes to

position 3. 4. SIN (positive if goes up first, negative if goes down first)

COS (positive if starts above middle line, negative if below), this sign goes at position 4

5. How many times the graph repeats itself in 2π, This number (2π / period) goes at position 5.

Note: If there is also sideways movement just look to see it it has moved right or left and how much.

1 2 3 4 5

Equation y = -2 sin (3x) + 1

Middle line

π− 2π−

2π π

23π π2

x

y

– 3

– 2

– 1

1

2

3

27

SPARE UNIT CIRCLES

0° or 2π COS x– 1 1

y

– 1

1

SIN 90° or

2π

180° or π

270° or 2

3π

0° or 2π COS x– 1 1

y

– 1

1

SIN 90° or

2π

180° or π

270° or 2

3π

28

CHAPTER 9 – DIFFERENTIATION

DIFFERENTIATION RULES The general way to find the gradient or )(xf ′ is to simplify

hxfhxf

hxf )()(

0lim

)( −+→

=′

eg Find )(xf ′ if f(x) = x2

222 2)()( hxhxhxhxf ++=+=+

h

hxhhh

xhxhxhh

xfhxfh

xf2222 2

0lim2

0lim)()(

0lim

)( +→

=−++

→=

−+→

=′

xhxh

xf 220

lim)( =+

→=′ Since h 0, 2x + h = 2x

This gives the following differentiation laws (gives )(xf ′ or dxdy )

1. naxy = 1−= nnaxdxdy

This means, multiply by old power and drop the power by 1.

eg 32xy = 22 6)2(3 xxdxdy

==

eg 45

2xy = 41

41

25)2(

45 xx

dxdy

==

eg 22 3

13

1 −== xx

y 333

32

32)

31(2

xxx

dxdy

−=−=−= −−

2. )()( xgxfy ±= )()( xgxfdxdy ′±′=

eg 325 23 −++= xxxy

1415 2 ++= xxdxdy

3. [ ]nxfy )(= [ ] )()( 1 xfxfndxdy n ′= −

This means if function in brackets, multiply by old power and differentiation of bracket, drop power of bracket by 1 eg 7)32(4 += xy

66 )32(56)32(4)2)(7( +=+= xxdxdy

eg 3232 )43(2

)43(2 −+=+

= xx

y

442

)43(36)43(2)6)(3(+

−=+−= −

xxxx

dxdy

29

4. )()( xgxfy = )()()()( xgxfxgxfdxdy ′+′=

If 2 different functions of x being multiplied together, differentiate first one and multiply by second, then add the second one differentiated multiplied by the first one. eg 35 )42(10 += xxy

2534

5234

)42(60)42(50

10)42(6)42(50

+++=

+++=

xxxxdxdy

xxxxdxdy

5. )()(

xgxfy =

[ ]2)()()()()(

xgxgxfxgxf

dxdy ′−′

=

eg 42

3

)15()42(

++

=xxy

24

323422

])15[()15(40)42()15()42(6

+++−++

=x

xxxxxdxdy

8

323422

)15()15()42(40)15()42(6

+++−++

=x

xxxxxdxdy

Could have also been written 423 )15()42( −++= xxy and differentiated using rule 4

GRAPHS OF )(xf )(xf ′ Special Point )(xf )(xf ′ X – intercepts Where f(x) = 0 - Maximum Turning Point

x-intercept Where 0)( =′ xf

Minimum Turning Point

- x-intercept Where 0)( =′ xf

Stationary Point of Inflection

- x-intercept Where 0)( =′ xf

Sloping Downwards - )(xf ′ is negative (below the x-axis) Sloping Upwards - )(xf ′ is positive (above the x-axis) General Shape - If )(xf was a cubic, )(xf ′ will be a quadratic

(one less power) etc Examples of General Shape

Cubic Quadratic

Linear Horizontal Line

Quadratic Linear

Horizontal line )(xf ′ = 0

x– 10 – 5 5 10

y

– 10

– 5

5

10

x– 10 – 5 5 10

y

– 10

– 5

5

10 f(x) f’(x) Turning Point x =2 X-Intercpt x =2 Slopes down x ≤ 2 Negative x ≤ 2 Slopes up x ≥ 2 Positive x ≥ 2

30

CHAPTER 10 – APPLICATIONS OF DIFFERENTIATION 1. EQUATION OF TANGENT

Use differentiation to find the gradient at the point (m). Then use the coordinates of the point to find the linear equation.

)( 11 xxmyy −=− Eg Find the equation of the tangent to the graph 723 −+= xxy when x = 4

23 2 += xdxdy when x = 4, 502)4(3 2 =+=

dxdy

The point is (4, 43 + 2(4) – 7) = (4, 65) )4(5065 −=− xy 6520050 +−= xy 13550 −= xy 2. EQUATION OF NORMAL

The normal is perpendicular (right angled) to the tangent. This means the gradient of the normal = -1 / gradient of tangent. Once you have worked this out use.

)( 11 xxmyy −=−

Eg Find the equation of the normal to the graph 723 −+= xxy when x = 4

23 2 += xdxdy when x = 4, 502)4(3 2 =+=

dxdy

Normal gradient = 02.050

1−=

−=

dxdy This is the only different step

The point is (4, 43 + 2(4) – 7) = (4, 65) )4(02.065 −−=− xy 6508.002.0 ++−= xy 08.6502.0 +−= xy 3. APRROXIMATE INCREASE

For the function )(xf , the increase of )(xf if x increases from a to a+h is given by the formula Approximate Increase / Change )(afh ′= Approximate New Value )()( afhaf ′+=

The Percentage Increase )(

)(100af

afh ′=

Eg Find the percentage increase for the function 75)( 2 += xxf as x goes from 4 to 4.01 Ans : a = 4 (original value), h = 0.01 (change)

xxf 10)( =′ 40)4(10)4()( ==′=′ faf 877)4(5)4()( 2 =+== faf

Approximate Change 4.0)40(01.0)( ==′= afh Approximate New Value 4.874.087)()( =+=′+= afhaf

(a, f(a))

( a+h, f(a)+hf’(a) )

h

hf’(a)

31

The Percentage Increase %46.08740

87)4.0(100

)()(100

===′

=af

afh

4. STATIONARY POINTS Wherever a function )(xf has stationary points (turning points, maximum point, minimum point, stationary point of inflection) this is where the gradient =0. Therefore at all stationary points 0)( =′ xf

5. TYPES OF STATIONARY POINTS Stationary Point

)(xf )(xf ′ Table Test )(xf ′′

Local Maximum (Turning Point)

0)( =′ xf )(xf ′ is positive just to left )(xf ′ is negative just to right

)(xf ′′ must also be negative

(frown) Local Minimum (Turning Point)

0)( =′ xf )(xf ′ is negative just to left )(xf ′ is positive just to right

)(xf ′′ must also be positive

(smile) Stationary Point of Inflection

0)( =′ xf )(xf ′ is same sign either side of point

)(xf ′′ =0

The table test table should look like this (GRADIENT TABLE)

x left Point eg x = 2 right

)(xf ′ + 0 -

Shape of f

Eg. Find the stationary points (and type for the equation) 1065.1)( 23 +−+= xxxxf

Ans For stationary points solve 0)( =′ xf 633)( 2 −+=′ xxxf 0633 2 =−+ xx 1=x and 2−=x These points are (1, 6.5) and (-2, 20) got by substituting into equation To determine the type of stationary points, find the second derivative 36)( +=′′ xxf Now substitute the stationary values into this to see if max or min 93)1(6)1( =+=′′f which is positive (smiling) = minimum turning point 93)2(6)2( −=+−=−′′f which is negative (frown) = maximum turning point 6. RATES OF CHANGE (Rate, Fast etc) The rate of change is given by the derivative function.

eg The rate of change of volume (V) with respect to time (t) is dtdV

eg The rate at which the Area is changing is dtdA

32

7. DISPLACEMENT (Movement)

Differentiate with respect to time Displacement Velocity Acceleration

y or )(xf .y or )(xf ′

..y or )(xf ′′

Integrate with respect to time

Initial means x = 0, so initial velocity is )0(f ′ etc

eg The displacement of a cars movement is given by 51043)( 23 +−+= xxxxf Calculate the initial acceleration Ans Displacement 51043)( 23 +−+= xxxxf Velocity 1089)( 2 −+=′ xxxf Acceleration 818)( +=′′ xxf Initial Acceleration 88)0(18)0( =+=′′f 8. RELATED RATES OF CHANGE

This is when you are asked to determine the rate at which something is changing, without the formula to differentiate it. This is when you need related rates of change. The steps are

Determine what you need to calculate Determine the rates given to you Use area / volume formulas to calculate other rates. Multiply relevant rates together

This is demonstrated by the following example. eg Water is filling a cylindrical tank at the rate of 20 cm3 / sec. at what rate is

the height increasing if the radius is 10 cm.

We need to work out dtdh but don’t have the formula between height and t

We are told that 20=dtdV

We have to work out how we can get that ??

×=dtdV

dtdh

We need a dh on the top and a dV to cancel the other one dVdh

dtdV

dtdh

×=

Knowing that hrV 2π= for a cylinder ππ 1002 == rdhdV

If π100=dhdV

π10011

==

dhdVdV

dh

33

ππ 51

100120 =×=×=

dVdh

dtdV

dtdh cm / sec

CHAPTER 11 – MORE DIFFERENTIATION OTHER RULES (Continued from Chapter 9)

6. )( xfey = )()( xfexfdxdy ′=

This means, differentiate the top bit and put it at the front, Keep everything else the same

eg xey 32= xx eedxdy 33 6)2(3 ==

eg 342 2

5 −+= xxey )5)(44( 342 2 −++= xxexdxdy

7. [ ])(log xfy e= )()(

xfxf

dxdy ′

=

This means, differentiate the bracket and put it at the top of the fraction, what was in the bracket goes at the bottom

eg )45(log 2 += xy e 45

102 +

=x

xdxdy

eg )3(log2 2xy e= xx

xdxdy 4)

36(2 2 ==

8. )](sin[ xfy = )](cos[)( xfxfdxdy ′=

This means, differentiate the bracket and put it at the front, change the sin to a cos.

eg )12sin(4 2 += xy

)12cos(16))12cos(4(4 22 +=+= xxxxdxdy

eg )4sin(3 xy −= )4cos(12))4cos(3(4 xxdxdy

−=−=

9. )](cos[ xfy = )](sin[)( xfxfdxdy ′−=

This means, differentiate the bracket and put it at the front, change the cos to a negative sin.

eg )15cos(3 += xy )15sin(15))15sin(3(5 +−=+−= xxdxdy

eg )6cos(5 2xy = )6sin(60))6sin(5(12 22 xxxxdxdy

−=−=

10. )](tan[ xfy = )]([sec)()](cos[)](cos[

)( 2 xfxfxfxf

xfdxdy ′=

′=

eg )5tan(4 xy = )5(cos

20)5(cos

)5(422 xxdx

dy==

eg )14tan(10 2 += xy )14(cos

80)14(cos

)8(102222 +

=+

=xx

xx

dxdy

34

You may also need to use the relationships 1)(sin)(cos 22 =+ xx )cos()sin()cos()sin()sin( ABBABA ±=± to simplify responses.

CHAPTER 12 – INTEGRATION (+c)

INTEGRATION RULES Opposite of differentiation. Always remember the + c (substituting any point on the function will let you find

this) You cannot integrate if 2 different functions of x are being multiplied, you would

need to expand the brackets before integrating. Eg ∫ +− dxxx )1)(2( must be expanded to ∫ −− dxxx )2( 2

Note : ∫ ′= dxxfxf )()( or dxdxdyy ∫=

Remember the correct symbols (dx) ∫ dxxf )( Remember to differentiate your final answer back to see if you are correct.

1. ∫ dxaxn = cnaxn

++

+

)1(

1

Basically this means raise the power by 1, divide by the new power.

eg. ∫ +=+= cxcxdxx 44

3

444

eg. cx

cxdxxdxx

+−=+−==∫ ∫ −−

51

51

51

51 12

2

eg. cxcxdxx +=+=∫ 25

25

23

54

2522

2. ∫ + dxbax n)( = cna

bax n

++

+ +

)1()( 1

Basically this means raise the power of bracket by 1, divide by the new power and also divide by differentiation of bracket

eg. ∫ ++

=++

=+ cxcxdxx20

)15()5(8

)15(2)15(288

7

eg. cxcxdxxdxx

++=++

=+=−∫ ∫

−126

21

)12(3)12(312

3 21

21

3. ∫′

dxxfxf)()( = [ ] cxf +)(ln

This rule is used when the original power of x is -1, obviously raising by 1 would make it 0, This is when the log law is needed. Put the bottom of fraction in a log, divide by differentiation of this.

eg. cxcx

dxxdxx e

e ++=++

=+=+∫ ∫ − )12(log2

2)12(log4

)12(412

4 1

35

eg.

cxcx

dxxdxx e

e ++=++

=+=+∫ ∫ − )110(log

21

10)110(log5

)110(5110

5 1

4. ∫ dxe xf )( = cxf

e xf

+′ )(

)(

Basically this means divide by differentiation of top part. Keep everything else the same.

eg. ∫ +=+

+ cedxex

x

255

1212

eg. ∫ +=+

+ cedxex

x

41010

8484

6. ∫ ++−=+ cakxk

dxakx )cos(1)sin(

Basically this means divide by differentiation of bracket. Change sin to a negative cos

eg. ∫ ++−

=+ cxdxx5

)25cos(6)25sin(6

eg. ∫ +−−

=− cxdxx4

)34cos()34sin(

7. ∫ ++=+ cakxk

dxakx )sin(1)cos(

Basically this means divide by differentiation of bracket. Change cos to a sin

eg. )2sin(2

)2sin(2)2cos(2 xcxdxx =+=∫

eg. ∫ +−

=− cxdxx5

)35sin()35cos(

DEFINITE INTEGRAL

∫b

a

dxxf )( is called the definite integral and does not need you to find the constant c.

This works out the area under the graph f(x) between [a,b] There are 5 properties of the definite integral (most make common sense when you think about areas under curves).

1. 0)( =∫a

a

dxxf

2. ∫∫∫ +=b

c

c

a

b

a

dxxfdxxfdxxf )()()( where c is in-between a and b

3. ∫∫ =b

a

b

a

dxxfkdxxkf )()( if all of the function is being multiplied by a

constant, it can be taken out and multiplied after integration

36

4. ∫∫∫ ±=±b

a

b

a

b

a

dxxgdxxfdxxgxf )()()()(

5. ∫∫ −=a

b

b

a

dxxfdxxf )()(

AREA UNDER A CURVE a) Using an approximation

In each the area between x=0 and x =4 is determined by dividing the region into rectangles of width = 1 unit. Obviously the lower the width the more rectangles and the more accuarate your approximation. The approximation on the left is called a left point estimate since the height of each rectangle is taken from the left side of the rectangle. The approximation on the right is called a right point estimate since the height of each rectangle is taken from the right side of the rectangle. Averaging this out makes the approximation more accurate.

b) Using integration to find the area exactly

∫b

a

dxxf )(

gives the exact or net signed area between the x-axis and the curve f(x) between x = a and x = b.

If parts of the area are below the x-axis then these parts give a negative area. Remember to change this area to a positive when adding it to other regions.

Therefore you need to graph the function first, then each region (if some regions above axis and some below) needs to done separately, and added together, so you will need to find x-intersects.

c) Using calculator Graph the function using y =, then graph 2nd, trace gives the Calc Menu, choose option 7. Enter the value of lower and upper limits.

x2 4 6 8

y

0.5

1

1.5

2

2.5

3

x2 4 6 8

y

0.5

1

1.5

2

2.5

3

37

EG. Determine the area between the x-axis, the line 62 −= xy , and the lines x= -1 and x = 6

There are 2 regions, one above the x-axis and one below. Work each out separately and make sure they are positive before adding them.

Between x=-1 and x = 3 (x-intercept)

[ ] 16)7(9)]1(6)1[()]3(63[6)62( 223

1

3

1

2 −=−−=−−−−−=−=−−−

∫ xxdxx

Between x=3 and x = 6

[ ] 9)9(0)]3(63[)]6(66[6)62( 226

3

6

3

2 =−−=−−−=−=−∫ xxdxx

The net signed area is 16+9 = 25 Check that this is correct using your calculator

x– 10 – 5 5 10

y

– 10

– 5

5

10

38

AREA BETWEEN CURVES Calculate The area of the region enclosed between the line 43 −= xy and the curve

62 +−= xy , and the lines x=-1 and x=4. ANS Between x=-1 and x=2 (the intersection), the parabola is above the line so calculate

∫−

−−+−2

1

2 )43(6 dxxx

Between x=2 and x=4, the line is above the parabola so calculate

∫ +−−−4

2

2 )6(43 dxxx

As can be seen you have to graph the functions first to calculate intersection points and which graph is above the other.

x– 10 – 5 5 10

y

– 10

– 5

5

10

39

CHAPTER 14 – DISCRETE RANDOM VARIABLES Number between 0 (impossible) and 1 (certain) which gives the chance of an event occurring. Chances of success Probability of something happening = Total number of chances The sample space is a list of all the possible outcomes eg for a coin the sample space is {head,tail}. TREE DIAGRAMS SHOWING PROBABILITY (Used when Dependant)

- Each branch or line has a probability written on it. - Sets of lines must add to 1. (ie 0.6 + 0.4 = 1, 0.5 + 0.5 = 1, 0.75 + 0.25 = 1 etc) -

Eg.

This is an example where two marbles are picked out of a bag, which contains 3 blue and 2 red marbles and the first marble is not replaced.

- Pr(each final possibility) = probability of all the branches leading there x together.

Eg. Pr(2 Blue) = Pr(Blue) then Pr(blue) = 0.6 x 0.5 = 0.3

- If final probability consists of a number of possibilities add each probability together. Eg. P(Blue and Red) = P(red then blue) or P(blue then red) = (0.4 x 0.75) + (0.6 x 0.5) = 0.6

NOTE: If you use the word OR, this means you add the probabilities, if you use the

word THEN you multiply the probabilities. Eg. Pr (1 Blue) = Blue then Red or Red then Blue = 0.6 x 0.5 + 0.4 x

0.75 GIVEN OR CONDITIONAL PROBABILITY

• Pr(A / B) means the probability of A happening given that B has already happened.

Blue

Red

0.6

0.4

Blue Red Blue Red

0.5 0.5 0.75 0.25

40

• Given removes many of the possibilities from your list or tree diagram giving you a new smaller sample space.

Pr(both occur)

• Pr (A given B) = Pr(given event occurs) Example of Given question, using previous example Given that the first marble was a blue, what is the probability of the second one being

blue?

Pr(both occur) Pr (second blue given first one was blue) = Pr(given event occurs)

Pr(blue, then blue) 0.6 x 0.5 Pr (second blue given first one was blue) = Pr(first one was blue) = 0.6 x 0.5 + 0.6 x 0.5 = 0.5 SOLVING WORD PROBLEMS INVOLVING PROBABILITY Think about the tree diagram or use the formulas below. Or Use the formulas below, with an example to show what they mean. 1. )Pr(1)Pr( AA −= the Complement rule

Pr(5 on a die) = 61 , so Pr(not a 5) =

611− =

65

2. )Pr(

)Pr()Pr(

)Pr()/Pr(Givenboth

ABAAB =

∩= the Conditional rule

Pr(5 / odd number on a die) = 31 =

)Pr(odd) and 5Pr(

odd2161

=

3. )/Pr()Pr()Pr( ABABA ×=∩ the Multiplication rule

Pr(drawing a king, then another king from a pack of cards)

= Pr(king) x Pr(king given that I have already got a king) = 524 x

513 =

265212

4. )Pr()Pr()Pr()Pr( BABABA ∩−+=∪ the Addition rule

P(red or king drawn from pack of cards)

= Pr(red) + Pr(king) – Pr(red and king) = 5228

522

524

5226

=−+

As you can see many of these are common sense, and you could have just drawn the relevant tree diagram to work out probabilities.

This is usually the second event. In this example Pr(B)

41

TYPES OF EVENTS 1. INDEPENDENT

First event happening doesn’t affect chances of next event eg. Two coins, if first one is a tail, this doesn’t affect the chances on the next coin

2. DEPENDANT

First event happening does affect chances of next event eg. Drawing an ace from a pack of cards, there is now only 51 cards and only three aces left so the first event now changes the chances on the second.

DISCRETE RANDOM VARIABLE Where the possible results are only integer values (eg. Tossing 3 coins, you can only get 0, 1, 2 or 3 tails, not 2.34 tails etc) DISCRETE PROBABILITY DISTRIBUTIONS (To Prove 2 things must occur)

• The probabilities must add up to 1 • Each probability must be between 0 and 1.

Probability distribution Eg. The results if three coins are tossed and “X” represents the number of tails.

x 0 1 2 3 Pr(X = x) 0.125 0.375 0.375 0.125

This means that Pr(getting 2 tails) = 0.375, Pr(getting 3 tails) = 0.125 This is a probability distribution since they add to 1. This can be shown as a graph. EXPECTED VALUE OF RANDOM VARIABLE X (MEAN)

∑ == )Pr(.)( xXxXE which is the sum of each possible outcome times by its probability This gives the long-run average value of X, also known as the mean (µ) In the example above 5.1)125.03()375.02()375.01()125.00()( =×+×+×+×=XE tails is the average result. EXPECTED VALUE OF A FUNCTION OF X

∑ == )Pr().()]([ xXxgxgE

In the example above 3)125.03()375.02()375.01()125.00()( 22222 =×+×+×+×=XE

0 1 2 3

0.375 0.25 0.125

x (Number of tails)

42

MEDIAN To determine the median work out at the value of X where the cumulative probability first gets to 0.5 or above. If it equals 0.5 exactly, make the median half way between X and the next value of X. If it is above 0.5, the median is this value of X. In the coins example

X 0 1 2 3 P(X ≤ x) 0.125 0.5 0.875 1

So the cumulative probability first reaches 0.5 or above when X=1, therefore the median is half way between here (X=1) and the next value (X=2), so the median = 1.5. MODE The mode is the value of X with the largest probability or chance of happening. In this case 1 or 2 tails (X=1 or 2) VARIANCE Measure of how spread out the values are from the mean

22 )]([)()( XEXEXVar −= Using the coins example

22 )]([)()( XEXEXVar −= 75.05.13)( 2 =−=XVar

A useful property of the variance is that Var(aX+b) = a2 Var(X) where a and b are constants STANDARD DEVIATION

)()( XVarXsd == σ . Usually 95% of the distribution lies within 2 standard deviations either side of the mean. In the coins example 866.075.0)( === σXsd

43

CHAPTER 15 – BINOMIAL DISTRIBUTION BERNOULLI SEQUENCES When a trial or event is repeated with only 2 possible outcomes (eg heads and tails) and the trials are independent (the result of one trial does not affect the chances of the next). MARKOV SEQUENCES When a trial or event is repeated with only 2 possible outcomes (eg heads and tails) and the trials are dependent (the result of one trial does affect the chances of the next). A tree diagram drawn with probabilities is useful to solve these problems. BINOMIAL PROBABILITY DISTRIBUTION (use when discrete, independent and given probability of success or can work it out) If an independent event (ie what happens doesn’t affect the chances of the next) occurs “n” times and the probability of success each time is “p”, then n = number of trials p = prob of success each time x = number of successes Pr(X = x) = n nCr x (p)x(1-p)n-x Note : nCr is in MATH PRB : 3 Eg. A die is rolled 7 times, what is the probability of getting a 4 three times.

X = “number of fours”, x = 3, p =61 , n = 7 times

Pr(X = 3) = 43 )65()

61(37nCr = 43 )

65()

61(35 = 0.07814

Obviously the maximum number of successes is “n” To do this on the Calculator. 2nd, VARS, then press enter on binompdf(n,p,x) To get a cumulative probability )( xXP ≤ 2nd, VARS, then press enter on binomcdf(n,p,x), this gives the probability that X is less than or equal to x. GRAPHS OF BINOMIAL DISTRIBUTIONS If the p(success each time) = 0.5 then the graph is symmetric, if above 0.5 it is more to the right (higher numbers) and they peak at the mean value. An example of a PDF (Probability distribution function) is below.

0 1 2 3

0.375 0.25 0.125

x (Number of tails)

44

EXPECTED VALUE (MEAN) AND VARIANCE The expected value npxE == µ)( Variance )1()( 2 pnpXVar −== σ Standard Deviation )1()( pnpXsd −== σ Again, Usually 95% of the distribution lies within 2 standard deviations either side of the mean. MINIMUM AMOUNT OF TRAILS NEEDED You are given the value of p, x and the )Pr( xX ≤ and you need to solve for n. Put the equation into the calculator (y=, enter equation, then scroll down the table until the probability is reached, 2nd graph) Eg. The probability of a shooter hitting a bullseye on any one shot is 0.2, what is the

smallest number of shots he should take to ensure a probability of more than 0.95 of hitting the bullseye at least once.

ANS We need to solve when )0Pr(1)1Pr( =−=≥ XX is first over 0.95 Into Y1 enter 1 – binompdf(x, 0.2, 0) Then go to the table to see when this first is above 0.95 (x=14) so a minimum of 14

trials are needed.

45

CHAPTER 16 – CONTINOUS RANDOM VARIABLES CONTINUOUS RANDOM VARIABLE Where the possible results can be any value, not just whole numbers (eg. Heights, weights etc). The probability that Pr(X=x) is always =0. For example the probability that someone is exactly 180 cm = 0 PROBABILITY DENSITY FUNCTION (PDF). A graph or equation )(xf showing the probability of a continuous variable. If continuous = density if discrete = distribution Eg.

• The area under the graph must equal 1 ∫∞

∞−

= 1)( dxxf

• Must all be above x-axis. 0)( ≥xf for all x The area under the graph shows probabilities. The actual probability at a point is actually 0 for continuous probability. Pr( a ≤ x ≤ b ) = Pr(between a and b) = Area under graph between x = a, and x = b. Use calculus (integrate between x = a and x = b) or just find the area of the relevant triangle or rectangle if a simple shape) Eg. If X is the random variable with probability density function F is given by:

05.0 x

elsewhere

x 20 ≤≤

Pr(X≥1) = [ ]∫ =−==2

1

2221

2 75.0)1(25.0)2(25.025.05.0 xxdx

MEAN The mean or expected value for a continuous random variable x is given by

∫∞

∞−

== dxxxfXE )(][ µ

Using the above example 25.0)5.0()( xxxxxf ==

0 1 2

1 0.5

x (Length)

46

E[X] = ∫∫ =−=⎥⎦⎤

⎢⎣⎡==

∞

∞−

2

0

332

0

32

68)0(

61)2(

61

35.05.0)( xdxxdxxxf

PERCENTILES AND MEDIAN The value (p) that (100q)% of the values lie below is calculated by solving the following

formula qdxxfp

=∫∞−

)(

For example the median (also known as 50th percentile) has 50% of the values below it so

the median is determined by solving 5.0)( =∫∞−

p

dxxf

Using the example

To find median solve ∫ =a

xdx0

5.05.0 [ ]∫ =−==a

a axxdx0

220

2 5.0)0(25.0)(25.025.05.0

So a2 = 2 a = 2 MODE The mode is the location of the highest point on the graph. This can be found by differentiating the function, then solving when it equals 0, to find the position of a turning point. Using the example, the highest part of the graph would be when x = 2 as there are no tuning points so Mode = 2. VARIANCE

Find ∫∞

∞−

= dxxfxXE )()( 22 then 222 )]([)()( XEXEXVar −== σ

Using the example 322 5.0)5.0()( xxxxfx ==

2)0(81)2(

81]

81[5.0)()( 442

04

2

0

322 =−==== ∫∫∞

∞−

xdxxdxxfxXE

Then 222.0682)]([)(

2222 =⎟

⎠⎞

⎜⎝⎛−=−= XEXEσ

STANDARD DEVIATION )(XVar=σ

Again, Usually 95% of the distribution lies within 2 standard deviations either side of the mean. In the example 471.0222.0)( === XVarσ RANGE The biggest value of x – the smallest value of x (ie the domain of the pdf). In the example 2 – 0 = 2 so the range is 2. INTERQUARTILE RANGE The value of the 75% percentile – the value of the 25% percentile. SPECIAL PROPERTIES For any random variable X bXaEbaXE +=+ )()( For any random variable X )()( 2 XVarabaXVar =+

The pdf of )(1)(a

bxfa

baXf −=+

For random variables X and Y )()()( YEXEYXE +=+ For random variables X and Y )()()( YVarXVarYXVar +=+

47

CHAPTER 17 – THE NORMAL DISTRIBUTION NORMAL OR GAUSSIAN DISTRIBUTIONS Normal distributions (or the bell curve) are symmetric and peak at the mean value. If you shade the area under the graph this gives probability, for example the probability of being within one standard deviation of the mean is always 0.68, so 68% are always within 1 sd of the mean in a normal distribution. There are always 95% within 2 standard deviations of the mean and 99.7% within 3 standard deviations of the mean (above or below). This is known as the 68-95-99.7% rule (must be remembered for non calculator exam) NOTE : This is a standard normal distribution since the horizontal axes has standard

deviations or Z scores, if there are normal scores on the horizontal axis this is a general normal distribution. The pdf of a standard (Z scores) normal distribution is the formula

2

21

21)(

xexf−

=π

The pdf of a general (X scores) normal distribution is the formula 2)(

21

21)( σ

µ

πσ

−−

=x

exf

This would need to be integrated to find areas under the graph, luckily there is a function on your calculator that does it for you.

The relationship between the 2 functions is

)Pr()Pr(σµ−

≤=≤aZaX

General Standard

σµ−a is known as the standardised or Z score, for example in a test if your

standardised score was +2, this means you received a mark 2 standard deviations above the mean (a very good score), a negative standardised score means you were below the mean.

3 sd below 2 sd below 1 sd below mean 1 sd above 2 sd above 3 sd above Z -3 -2 -1 0 1 2 3

This area is 0.68 or 68%

48

USE OF CALCULATOR 1. Pr(a ≤ X ≤ b) = Pr(X is between a and b)

= normalcdf(Za,Zb)

eg. A class averaged 62 for a test with a standard deviation of 14. What is the probability of getting a score less than 34.

−∞=a 34=b

−∞=−∞−

=14

62aZ Put in calculator as -10^99

214

6234−=

−=bZ

Pr( ∞− ≤ X ≤ 34) = Pr( ∞− ≤ Z ≤ -2) = normalcdf(-10^99,-2) = 0.02275 NOTE : To do this in a non-calculator exam remembering the 68-95-99.7% rule and

the fact that the graph is symmetric says that 95% of scores lie within 2 standard deviations of the mean. Therefore 5% or 0.05 are either 2 standard deviations above or below the mean. Since it’s symmetric 2.5% (or approx 0.025) must be less than 2 standard deviations below the mean.

2. Or more simply you can do Pr(a ≤ X ≤ b) = Pr(X is between a and b)

= normalcdf(a,b,µ,σ)

eg. A class averaged 62 for a test with a standard deviation of 14. What is the probability of getting a score less than 34.

−∞=a 34=b , µ = 62, σ = 14 Pr( ∞− ≤ X ≤ 34) = normalcdf(-10^99,34,62,14) = 0.02275

3. To solve Pr(Z ≤ c) = p then c = invNorm(p) This obviously gives the Z value to convert to an X-value use the formula

µσ += ZX

Eg. A class averaged 62 for a test with a standard deviation of 14. What score did 80% of the class beat.

p = 0.8 Pr(Z ≤ c) = 0.8, then c = invNorm(0.8) = 0.8416 78.7362148416.0 =+×=X

4. Or more simply you can do

To solve Pr(X ≤ c) = p then c = invNorm(p,µ,σ)

Eg. A class averaged 62 for a test with a standard deviation of 14. What score did 80% of the class beat.

p = 0.8 Pr(X ≤ c) = 0.8, then c = invNorm(0.8,62,14) = 73.78

Note : If you are asked for the interval that 60% of the scores are between ie Pr(a ≤ X ≤ b) = 0.6, use the fact that the graph is symmetric to determine that 20% of the scores must be below a, and 20% of the scores are above b (or 80% are below b). Therefore you solve Pr(X ≤ a) = 0.2 or invNorm(0.2,62,14)

2nd, vars, highlight and enter Must calculate the standardised scores.

3

PROBABILITY

DISCRETE Only whole numbers

CONTINUOUS Can be any number

DEPENDENT One trial affects the chances of it

happening next time

INDEPENDENT One trial doesn’t affect the next

BINOMIAL THEOREM n = number of trials

p = prob of success each time x = number of successes

Pr(X = x) = n nCr x (p)x(1-p)n-x

or

Pr(X = x) = binompdf(n,p,x)

Note : nCr is in MATH PRB : 3 Binompdf is in DISTR (2nd VARS)

To get Probability that it is less than

or equal to x ( )Pr( xX ≤ ) use binomcdf(n,p,x)

Mean np=µ

Variance )1(2 pnp −=σ

Standard Dev = var=σ

Graph peaks at the mean

GIVEN

)Pr()Pr()Pr(

BBABA ∩

= or

P (A happening given that B has already happened) = Pr(both happening) ÷ P(B)

TREE DIAGRAM

TABLES Each probability has to be between 0

and 1, and they all must add to 1

X 0 1 2 Pr(X = x) 0.25 0.5 0.25

Mean E(X) =∑ = )Pr( xXx

=sum of each possible outcome times by its probability

(eg 0 x 0.25 + 1 x 0.5 + 2 x 0.25) = 1

Variance E(X2) – [E(X)]2 Where E(X2) = ∑ = )Pr(2 xXx

(eg E(X2) = 02 x 0.25 + 12 x 0.5 + 22 x 0.25 = 1.5, then

Var = E(X2) – [E(X)]2 = 1.5 – (1)2 = 0.5

Mode = Most likely = 1 in this case

Median = when )Pr( xX ≤ is first over 0.5. In this case Median = 1

NORMAL

68-95-99.7% rule

Standardised Score Z score = (score – mean) ÷

standard deviation

Probability between 2 Standardised (Z scores)

)Pr( bZa ≤≤ =

normalcdf(Za,Zb)

Probability between 2 “normal scores”

)Pr( bXa ≤≤ =

normalcdf(a,b,µ,σ)

To solve Pr(Z ≤ c) = p c = invNorm(p)

To solve Pr(X ≤ c) = p

c = invNorm(p,µ,σ)

GIVEN A PROBABILITY DENSITY FUNCTION

Graph must be all above the x axis, the area under the graph

must equal 1 and the probabilities are worked out by

calculating areas under the graph (integrate)

Mean ∫∞

∞−

= dxxxfXE )()(

Variance E(X2) – [E(X)]2

Where ∫∞

∞−

= dxxfxXE )()( 22

Mode = highest part of graph (differentiate, solve = 0 to get

turning point)

Median = solve for a

5.0)( =∫∞−

a

dxxf