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Higher Unit 1Higher Unit 1
www.mathsrevision.comwww.mathsrevision.com
Finding the gradient for a polynomialDifferentiating a polynomial
Differentiating Negative Indices Differentiating Roots
Differentiating Brackets
Differentiating Fraction TermsDifferentiating with Leibniz Notation
Exam Type Questions
Equation of a Tangent Line
Increasing / Decreasing functionsMax / Min and Inflection Points
Curve Sketching
Max & Min Values on closed IntervalsOptimization
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On a straight line the gradient remains constant, however with curves the gradient changes continually, and the gradient at any point is in fact the same as the gradient of the tangent at that point.
The sides of the half-pipe are very steep(S) but it is not very steep near the base(B).
B
S
Gradients & CurvesHigher Outcome 3
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A
Gradient of tangent = gradient of curve at A
B
Gradient of tangent = gradient of curve at B
Gradients & CurvesHigher Outcome 3
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Gradients & Curves
To find the gradient at any point on a curve we need
to modify the gradient formula
Higher Outcome 3
For the function y = f(x) we do this by taking the point (x, f(x))
and another “very close point” ((x+h), f(x+h)).Then we find the gradient between the two.
(x, f(x))
((x+h), f(x+h))
True gradient
Approx gradient
2 1
2 1
--
y ym
x x
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The gradient is not exactly the same but is
quite close to the actual value We can improve the approximation by making the value of h
smallerThis means the two points are closer together.
(x, f(x))
((x+h), f(x+h))
True gradient
Approx gradient
Gradients & CurvesHigher Outcome 3
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We can improve upon this approximation by making the value of h even smaller.
(x, f(x))
((x+h), f(x+h))True gradientApprox gradient
So the points are even closer together.
Gradients & CurvesHigher Outcome 3
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Higher Outcome 3
Derivative
We have seen that on curves the gradient changes continually and is dependant on the position on
the curve. ie the x-value of the given point.
We can use the formula for the curve to produce a formula for the gradient.
This process is called DIFFERENTIATING
or FINDING THE DERIVATIVE
Differentiating
Finding the GRADIENT
Finding the rate of change
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If the formula/equation of the curve is given by f(x)Then the derivative is called f '(x) - “f dash x”
There is a simple way
of finding f '(x) from f(x).
f(x) f '(x)
2x2 4x 4x2 8x 5x10 50x9
6x7 42x6
x3 3x2
x5 5x4
x99 99x98
DerivativeHigher Outcome 3
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Rules for Differentiating
These can be given by the simple flow diagram ...
multiply by the power
reduce the power by 1
Or
If f(x) = axn
then f '(x) = naxn-1
NB: the following terms & expressions mean the same
GRADIENT, DERIVATIVE, RATE OF CHANGE, f '(x)
Derivative
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Example 1 A curve has equation f(x) = 3x4
Its gradient is f '(x) = 12x3
At the point where x = 2 the gradient isf '(2) = 12 X 23 = 12 X 8 = 96
Example 2 A curve has equation f(x) = 3x2
Find the formula for its gradient and find the gradient when x = -4 .Its gradient is f '(x) = 6x
At the point where x = -4 the gradient isf '(-4) = 6 X -4 =-24
DerivativeHigher Outcome 3
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g '(x) = 20x3 - 20x4 g '(2) = 20 X 23 - 20 X 24
= 160 - 320
= -160
DerivativeHigher Outcome 3
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Your turn
Page 91 ex 6D ( odd numbers ) Page 92-93 ex. 6E ( even
numbers )
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Special Points
(I) f(x) = ax, where a is any real no.
If f(x) = ax = ax1
then f '(x) = 1 X ax0
= a X 1 = a
Index Laws
x0 = 1
So if g(x) = 12x then g '(x) = 12
Also using y = mx + c
The line y = 12x has gradient 12,
and derivative = gradient !!
Higher Outcome 3
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If f(x) = a = a X 1 = ax0
then f '(x) = 0 X ax--1= 0
Index Laws
x0 = 1
So if g(x) = -2 then g '(x) = 0
Also using formula y = c , (see outcome1 !)
The line y = -2 is horizontal so has gradient 0 !
Special PointsHigher Outcome 3
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Example 6
h(x) = 5x2 - 3x + 19
so h '(x) = 10x - 3
and h '(-4) = 10 X (-4) - 3
= -40 - 3 = -43
Example 7
k(x) = 5x4 - 2x3 + 19x - 8, find k '(10) .
k '(x) = 20x3 - 6x2 + 19
So k '(10) = 20 X 1000 - 6 X 100 + 19
= 19419
DerivativeHigher Outcome 3
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Find the points on the curve
f(x) = x3 - 3x2 + 2x + 7 where the gradient is 2.
NB: gradient = derivative = f '(x)
We need f '(x) = 2
ie 3x2 - 6x + 2 = 2
or 3x2 - 6x = 0
ie 3x(x - 2) = 0
ie 3x = 0 or x - 2 = 0
so x = 0 or x = 2
Now using original formula
f(0) = 7
f(2) = 8 -12 + 4 + 7
= 7
Points are (0,7) & (2,7)
DerivativeHigher Outcome 3
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Negative Indices
Index Law x-m = 1 xm
Consider ….. am X a-m = am+(-m) =a0 = 1
also am X 1 = am
1
This gives us the following
NB: Before we can differentiate a term it must be in the form axn .
Bottom line terms get negative powers !!
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f(x) = 1 = x2
x -2
So f '(x) = -2x-3 =
-2 x3
Example 10
g(x) = -3 x4
= -3x-4
So g '(x) = 12x-5 =
12
x5
Negative IndicesHigher Outcome 3
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Example 11
h(y) = 4 3y3
= 4/3 y-3
So h '(y) =
-12/3 y-4 = -4 y4
Example 12
k(t) = 3 4t2/3
= 3/4 t-2/3
So k '(t) =- -6/12 t-5/3 = -1 2t5/3
Negative IndicesHigher Outcome 3
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Example 13
The equation of a curve is f(x) = 8 (x 0) x4
Find the gradient at the point where x = -2 .
f(x) = 8 x4
= 8x-4 so f '(x) = -32x-5 =
-32 x5
Required gradient = f '(-2) =-32 (-2)5
= -32 -32
= 1
Negative IndicesHigher Outcome 3
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Differentiating Roots
Fractional Indices
x X x = x
x1/2 X x1/2 = x1/2+1/2 = x1 = x
So it follows that …..
Similarly 3x = x1/3 and 4x = x1/3
x = x1/2
In general nx = x1/n
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Also nxm or (nx)m = xm/n
Check the following using the power button on your calc.
641/2 = 64 = 8
641/3 = 364 = 4
25-1/2 = 1 = 1 or 0.2
25 5
163/4 = (416)3 = 23 = 8
1255/3 = (3125)5 = 55 = 3125
Fractional Powers
top line - power
bottom line - root
Higher Outcome 3
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so h'(y) = -7/3 y-10/3 = -7 33y10
Example 14
f(x) = x = x1/2 so f '(x) = 1/2 x-1/2 = 1 2x
Example 15
= t5/2 so g'(t) = 5/2 t3/2 = 5t3
2Example 16
h(y) = 1 3y7
= y-7/3
g(t) = t5
Differentiating RootsHigher Outcome 3
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Example 17Find the rate of change at the point where x = 4 on the curve with equation g(x) = 4 .
x g(x) = 4
x
= 4x-1/2
NB: rate of change = gradient = g'(x) .
g'(x) = -2x-3/2 = -2
(x)3
so g'(4) = -2 (4)3
= -2/8 = -1/4 .
Differentiating RootsHigher Outcome 3
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Brackets
Basic Rule: Break brackets before you differentiate !
Example 18 h(x) = 2x(x + 3)(x -3)
= 2x(x2 - 9)
= 2x3 - 18x
So h'(x) = 6x2 -18
Also h'(-2) = 6 X (-2)2 -18= 24 - 18 = 6
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Recall 17 7 7
Fractions
Reversing the above we get the following “rule” !
This can be used as follows …..
a + bc
a bc c
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Example 20
f(x) = 3x3 - x + 2 x2
= 3x - x-1 + 2x-2
f '(x) = 3 + x-2 - 4x-3
Fractions
= 3x3 - x + 2 x2 x2 x2
= 3 + 1 - 4 x2 x3
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Example 21 (tricky)
g(y) = (y + y)(y + 1) yy
= (y + y1/2)(y + 1)
y X y1/2
= y2 + y3/2 + y + y1/2
y3/2
= y2 + y3/2 + y + y1/2
= y1/2 + 1 + y-1/2 + y-1
y3/2 y3/2 y3/2 y3/2
Change to powers √=½
Brackets
Single fractions
Correct form
Indices
FractionsHigher Outcome 3
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= 1/4 - 1/16 - 1/16 = 1/8
= 1 - 1 - 1 2y1/2 2y3/2 y2
Also g'(4) = 1 - 1 - 1 2 X 41/2 2 X 43/2 42
= 1 - 1 - 1 2 X 4 2 X (4)3 16
FractionsHigher Outcome 3
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If y is expressed in terms of x then the derivative is written as dy/dx .
Leibniz Notation
Leibniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc.
eg y = 3x2 - 7x so dy/dx = 6x - 7 .
Example 22 Find dQ/dR
NB: Q = 9R2 - 15R-3
So dQ/dR = 18R + 45R-4 = 18R + 45 R4
Q = 9R2 - 15 R3
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Example 23
A curve has equation y = 5x3 - 4x2 + 7 .
Find the gradient where x = -2 ( differentiate ! )
gradient = dy/dx = 15x2 - 8x
if x = -2 then
gradient = 15 X (-2)2 - 8 X (-2)
= 60 - (-16) = 76
Leibniz NotationHigher Outcome 3
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Newton’s 2ndLaw of Motion
s = ut + 1/2at2 where s = distance & t = time.
Finding ds/dt means “diff in dist” “diff in time”
ie speed or velocity
so ds/dt = u + at
but ds/dt = v so we get v = u + at
and this is Newton’s 1st Law of Motion
Real Life ExamplePhysics
Higher Outcome 3
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y = mx +c
y = f(x)
Equation of Tangents
tangent
NB: at A(a, b) gradient of line = gradient of curve
gradient of line = m (from y = mx + c )
gradient of curve at (a, b) = f (a)
it follows that m = f (a)
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Higher Outcome 3
Equation of Tangents
Example 24
Find the equation of the tangent to the curve
y = x3 - 2x + 1 at the point where x = -1.
Point: if x = -1 then y = (-1)3 - (2 X -1) + 1
= -1 - (-2) + 1= 2 point is (-1,2)
Gradient: dy/dx = 3x2 - 2
when x = -1 dy/dx = 3 X (-1)2 - 2
= 3 - 2 = 1 m = 1
Tangent is line so we need a point plus the gradient then we can use the formula y - b =
m(x - a) .
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we get y - 2 = 1( x + 1)
or y - 2 = x + 1
or y = x + 3
point is (-1,2)
m = 1
Equation of Tangents
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Example 25
Find the equation of the tangent to the curve y = 4 x2
at the point where x = -2. (x 0)
Also find where the tangent cuts the X-axis and Y-axis.Point:when x = -2 then y = 4
(-2)2 = 4/4 =
1
point is (-2, 1)
Gradient: y = 4x-2 so dy/dx = -8x-3 = -8 x3
when x = -2 then dy/dx = -8 (-2)3
= -8/-8 = 1m =
1
Equation of TangentsHigher Outcome 3
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Now using y - b = m(x - a)
we get y - 1 = 1( x + 2)
or y - 1 = x + 2
or y = x + 3
Axes Tangent cuts Y-axis when x = 0
so y = 0 + 3 = 3
at point (0, 3)
Tangent cuts X-axis when y = 0
so 0 = x + 3 or x = -3 at point (-3, 0)
Equation of TangentsHigher Outcome 3
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Example 26 - (other way round)
Find the point on the curve y = x2 - 6x + 5 where the gradient of the tangent is 14.
gradient of tangent = gradient of curve
dy/dx = 2x - 6
so 2x - 6 = 14
2x = 20 x = 10
Put x = 10 into y = x2 - 6x + 5
Giving y = 100 - 60 + 5 = 45 Point is (10,45)
Equation of TangentsHigher Outcome 3
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Increasing & Decreasing Functions and Stationary Points
Consider the following graph of y = f(x) …..
X
y = f(x)
a b c d e f+
+
+
++
-
-
0
0
0
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In the graph of y = f(x)
The function is increasing if the gradient is positive
i.e. f (x) > 0 when x < b or d < x < f or x > f . The function is decreasing if the gradient is negativeand f (x) < 0 when b < x < d .
The function is stationary if the gradient is zeroand f (x) = 0 when x = b or x = d or x = f .These are called STATIONARY POINTS.
At x = a, x = c and x = e
the curve is simply crossing the X-axis.
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
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Example 27
For the function f(x) = 4x2 - 24x + 19 determine the intervals when the function is decreasing and
increasing.f (x) = 8x - 24
f(x) decreasing when f (x) < 0 so 8x - 24 < 0 8x < 24
x < 3
f(x) increasing when f (x) > 0 so 8x - 24 > 0
8x > 24x > 3
Check: f (2) = 8 X 2 – 24 = -8
Check: f (4) = 8 X 4 - 24= 8
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
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Example 28
For the curve y = 6x – 5/x2 Determine if it is increasing or decreasing when x = 10.
= 6x - 5x-2
so dy/dx = 6 + 10x-3
when x = 10 dy/dx = 6 + 10/1000
= 6.01
Since dy/dx > 0 then the function is increasing.
Increasing & Decreasing Functions and Stationary Points
y = 6x - 5 x2
= 6 + 10 x3
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Example 29Show that the function g(x) = 1/3x3 -3x2 + 9x -10
is never decreasing.
g (x) = x2 - 6x + 9
= (x - 3)(x - 3)= (x - 3)2
Since (x - 3)2 0 for all values of x
then g (x) can never be negative
so the function is never decreasing.
Squaring a negative or a positive value produces a positive value, while 02 = 0. So you will never
obtain a negative by squaring any real number.
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
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Example 30
Determine the intervals when the function
f(x) = 2x3 + 3x2 - 36x + 41
is (a) Stationary (b) Increasing (c) Decreasing.
f (x) = 6x2 + 6x - 36
= 6(x2 + x - 6)
= 6(x + 3)(x - 2)
Function is stationary when f (x) = 0
ie 6(x + 3)(x - 2) = 0 ie x = -3 or x = 2
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
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determine when f (x) is positive & negative.
x -3 2
f’(x) + 0 - 0 +
Function increasing when f (x) > 0
ie x < -3 or x > 2
Function decreasing when f (x) < 0
ie -3 < x < 2
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
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Stationary Points and Their Nature
Consider this graph of y = f(x) again
X
y = f(x)
a b c+
+
+
+
+
-
-
0
0
0
Higher Outcome 3
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When x = a we have a maximum turning point (max TP)When x = b we have a minimum turning point (min TP)When x = c we have a point of inflexion (PI)
Each type of stationary point is determined by the gradient ( f(x) ) at either side of the stationary
value.
Stationary Points and Their Nature
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Maximum Turning point
x af(x) + 0 -
Minimum Turning Point
x bf(x) - 0
+
Stationary Points and Their Nature
Higher Outcome 3
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of Inflection
x c
f(x) + 0 +
Falling Point of Inflection
x d
f(x) - 0 -
Stationary Points and Their Nature
Higher Outcome 3
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Example 31Find the co-ordinates of the stationary point on the curve y = 4x3 + 1 and determine its nature.
SP occurs when dy/dx = 0so 12x2 = 0
x2 = 0
x = 0
Using y = 4x3 + 1
if x = 0 then y = 1
SP is at (0,1)
Stationary Points and Their Nature
Higher Outcome 3
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x 0
dy/dx + 0 +
So (0,1) is a rising point of inflexion.
Stationary Points and Their Nature
dy/dx = 12x2
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Example 32
Find the co-ordinates of the stationary points on the curve y = 3x4 - 16x3 + 24 and determine their nature.
SP occurs when dy/dx = 0
So 12x3 - 48x2 = 0
12x2(x - 4) = 0
12x2 = 0 or (x - 4) = 0
x = 0 or x = 4
Using y = 3x4 - 16x3 + 24
if x = 0 then y = 24
if x = 4 then y = -232
SPs at (0,24) & (4,-232)
Stationary Points and Their Nature
Higher Outcome 3
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Nature Table
x 0 4
dy/dx - 0 - 0 +
So (0,24) is a Point of Infection
and (4,-232) is a minimum Turning Point
Stationary Points and Their Nature
dy/dx=12x3 - 48x2
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Example 33Find the co-ordinates of the stationary points on the curve y = 1/2x4 - 4x2 + 2 and determine their nature.
SP occurs when dy/dx = 0
So 2x3 - 8x = 0
2x(x2 - 4) = 0
2x(x + 2)(x - 2) = 0
x = 0 or x = -2 or x = 2
Using y = 1/2x4 - 4x2 + 2if x = 0 then y = 2
if x = -2 then y = -6
SP’s at(-2,-6), (0,2) & (2,-6)
if x = 2 then y = -6
Stationary Points and Their Nature
Higher Outcome 3
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Nature Table
x 0
dy/dx
-2 2
- 0 + 0 - 0 +
So (-2,-6) and (2,-6) are Minimum Turning Points
and (0,2) is a Maximum Turning Points
Stationary Points and Their Nature
Higher Outcome 3
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Curve Sketching
Note: A sketch is a rough drawing which includes important details. It is not an accurate scale
drawing.Process
(a) Find where the curve cuts the co-ordinate axes. for Y-axis put x = 0
for X-axis put y = 0 then solve.(b) Find the stationary points & determine their
nature as done in previous section.
(c) Check what happens as x +/- .This comes automatically if (a) & (b) are correct.
Higher Outcome 3
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Dominant Terms
Suppose that f(x) = -2x3 + 6x2 + 56x - 99
As x +/- (ie for large positive/negative values) The formula is approximately the same as f(x) = -2x3
As x + then y -
As x - then y +
Graph roughly
Curve SketchingHigher Outcome 3
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Example 34
Sketch the graph of y = -3x2 + 12x + 15(a) Axes If x = 0 then y = 15
If y = 0 then -3x2 + 12x + 15 = 0
( -3)
x2 - 4x - 5 = 0
(x + 1)(x - 5) = 0
x = -1 or x = 5
Graph cuts axes at (0,15) , (-1,0) and (5,0)
Curve SketchingHigher Outcome 3
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(b) Stationary Points occur where dy/dx = 0
so -6x + 12 = 0
6x = 12
x = 2
If x = 2
then y = -12 + 24 + 15 = 27
Nature Table
x 2
dy/dx + 0 -So (2,27)
is a Maximum Turning Point
Stationary Point is (2,27)
Curve SketchingHigher Outcome 3
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using y = -3x2
as x + then y -
as x - then y -
Sketching
X
Y
y = -3x2 + 12x + 15
Curve Sketching
Cuts x-axis at -1 and 5
Summarising
Cuts y-axis at 15-1 5
Max TP (2,27)(2,27)
15
Higher Outcome 3
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Example 35Sketch the graph of y = -2x2 (x - 4)(a) Axes
If x = 0 then y = 0 X -4 = 0If y = 0 then -2x2 (x - 4) = 0
x = 0 or x = 4
Graph cuts axes at (0,0) and (4,0) .
-2x2 = 0 or (x - 4) = 0
(b) SPs
y = -2x2 (x - 4) = -2x3 + 8x2
SPs occur where dy/dx = 0
so -6x2 + 16x = 0
Curve SketchingHigher Outcome 3
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-2x(3x - 8) = 0
-2x = 0 or (3x - 8) = 0
x = 0 or x = 8/3
If x = 0 then y = 0 (see part (a) ) If x = 8/3 then y = -2 X (8/3)2 X (8/3 -4) =512/27
naturex 0 8/3
dy/dx - 0 + 0 -
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(c) Large valuesusing y = -2x3 as x + then y -
as x - then y +
Sketch
Xy = -2x2 (x – 4)
Curve Sketching
Cuts x – axis at 0 and 40 4
Max TP’s at (8/3, 512/27) (8/3, 512/27)
Y
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Example 36
Sketch the graph of y = 8 + 2x2 - x4
(a) Axes If x = 0 then y = 8 (0,8)If y = 0 then 8 + 2x2 - x4 = 0
Graph cuts axes at (0,8) , (-2,0) and (2,0)
Let u = x2 so u2 = x4
Equation is now 8 + 2u - u2 = 0
(4 - u)(2 + u) = 0
(4 - x2)(2 + x2) = 0
or (2 + x) (2 - x)(2 + x2) = 0So x = -2 or x = 2 but x2 -2
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(b) SPs SPs occur where dy/dx = 0
So 4x - 4x3 = 0 4x(1 - x2) = 0
4x(1 - x)(1 + x) = 0 x = 0 or x =1 or x = -1
Using y = 8 + 2x2 - x4
when x = 0 then y = 8
when x = -1 then y = 8 + 2 - 1 = 9 (-1,9) when x = 1 then y = 8 + 2 - 1 = 9
(1,9)
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x -1 0 1
dy/dx + 0 - 0 + 0 -
So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .
Curve SketchingHigher Outcome 3
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Cuts y – axis at 8Cuts x – axis at -2 and 2
(c) Large valuesUsing y = -
x4as x + then y -
as x - then y -
Sketch is
X
Y
-2 28
(-1,9) (1,9)
y = 8 + 2x2 - x4
Max TP’s at (-1,9) (1,9)
Curve Sketching
Summarising
Higher Outcome 3
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Max & Min on Closed Intervals
In the previous section on curve sketching we dealt with the entire graph.
In this section we shall concentrate on the important details to be found in a small section of
graph.Suppose we consider any graph between the points
where x = a and x = b (i.e. a x b)
then the following graphs illustrate where we would expect to find the maximum & minimum values.
Higher Outcome 3
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y =f(x)
Xa b
(a, f(a))
(b, f(b)) max = f(b) end point
min = f(a) end point
Max & Min on Closed Intervals
Higher Outcome 3
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x
y =f(x)(b, f(b))
(a, f(a))
max = f(c ) max TP
min = f(a) end point
a b
(c, f(c))
c NB: a < c < b
Max & Min on Closed Intervals
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y =f(x)
xa b
c
(a, f(a))
(b, f(b))
(c, f(c))
max = f(b) end point
min = f(c) min TP
NB: a < c < b
Max & Min on Closed Intervals
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From the previous three diagrams we should be able to see that the maximum and minimum values of f(x) on the closed interval a x b can be found either at the end points or at a stationary point between the two end points
Example 37
Find the max & min values of y = 2x3 - 9x2 in the interval where -1 x 2.
End points If x = -1 then y = -2 - 9 = -11
If x = 2 then y = 16 - 36 = -20
Max & Min on Closed Intervals
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Stationary pointsdy/dx = 6x2 - 18x = 6x(x - 3)
SPs occur where dy/dx = 0
6x(x - 3) = 0
6x = 0 or x - 3 = 0
x = 0 or x = 3
in interval not in interval
If x = 0 then y = 0 - 0 = 0
Hence for -1 x 2 , max = 0 & min = -20
Max & Min on Closed Intervals
Higher Outcome 3
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Using function notation we can say that
Domain = {xR: -1 x 2 }
Range = {yR: -20 y 0 }
Max & Min on Closed Intervals
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Optimization
Note: Optimum basically means the best possible.
In commerce or industry production costs and profits can often be given by a mathematical
formula.
Optimum profit is as high as possible so we would look for a max value or max TP.
Optimum production cost is as low as possible so we would look for a min value or min TP.
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Higher Outcome 3Example 41
OptimizationQ. What is the maximum
volume
We can have for the given dimensions
A rectangular sheet of foil measuring 16cm X 10 cm has four small squares cut from each corner.
16cm
10cmx cm
NB: x > 0 but 2x < 10 or x < 5ie 0 < x < 5
This gives us a particular interval to consider !
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(16 - 2x) cm(10 - 2x) cm
x cm
The volume is now determined by the value of x so we can write
V(x) = x(16 - 2x)(10 - 2x)
= x(160 - 52x + 4x2)
= 4x3 - 52x2 +160x
We now try to maximize V(x) between 0 and 5
Optimization
By folding up the four flaps we get a small cuboid
Higher Outcome 3
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5End Points
V(0) = 0 X 16 X 10 = 0
V(5) = 5 X 6 X 0 = 0
SPs V '(x) = 12x2 - 104x + 160
= 4(3x2 - 26x + 40)
= 4(3x - 20)(x - 2)
OptimizationHigher Outcome 3
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ie 4(3x - 20)(x - 2) = 0
3x - 20 = 0 or x - 2 = 0
ie x = 20/3 or x = 2
not in intervalin interval
When x = 2 then
V(2) = 2 X 12 X 6 = 144
We now check gradient near x = 2
OptimizationHigher Outcome 3
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x 2
V '(x) + 0 -
Hence max TP when x = 2
So max possible volume = 144cm3
Nature
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Example 42
When a company launches a new product its share of the market after x months is calculated by the formula
So after 5 months the share is
S(5) = 2/5 – 4/25 = 6/25
Find the maximum share of the market
that the company can achieve.
(x 2)2
2 4( )S x
x x
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End points S(2) = 1 – 1 = 0
There is no upper limit but as x S(x) 0.
SPs occur where S (x) = 0
3 2
8 2'( ) 0S x
x x
1 22
2 4( ) 2 4 S x x x
x x
1 2 2 32 3
2 8'( ) 2 4 2 8S x x x x x
x x
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8x2 = 2x3
8x2 - 2x3 = 0
2x2(4 – x) = 0
x = 0 or x = 4
Out with interval In interval
We now check the gradients either side of 4
3 2
8 2'( ) 0S x
x xrearrange
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x 4
S (x) + 0 -
S (3.9 ) = 0.00337…
S (4.1) = -0.0029…
Hence max TP at x = 4
And max share of market = S(4) = 2/4 – 4/16
= 1/2 – 1/4
= 1/4
Optimization
Nature
Higher Outcome 3
Differentiation
Higher Mathematics
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