Absolute Value in Algebra

Absolute Value means ...

... how far a number is from zero:

"6" is 6 away from zero,

and "−6" is also 6 away from zero.

So the absolute value of 6 is 6,

and the absolute value of −6 is also 6

Absolute Value Symbol

To show you want the absolute value of something, you put "|" marks either

side (called "bars"), like these examples:

|−5| = 5 |7| = 7

The "|" can be found just above the enter key on most keyboards.

More Formal

So, when a number is positive or zero we leave it alone, when it is negative we

change it to positive.

This can all be written like this:

Which says the absolute value of x equals:

x when x is greater than zero

0 when x equals 0

−x when x is less than zero (this "flips" the number back to positive)

Example: what is |−17| ?

Well, it is less than zero, so we need to calculate "−x":

− ( −17 ) = 17

(Because two minuses make a plus)

Useful Properties

Here are some properties of absolute values that can be useful:

|a| ≥ 0 always!

That makes sense ... |a| can never be less than zero.

|a| = √(a2)

Squaring a makes it positive or zero (for a as a Real Number). Then

taking the square root will "undo" the squaring, but leave it positive or

zero.

|a × b| = |a| × |b|

Means these are the same:

the absolute value of (a times b), and

(the absolute value of a) times (the absolute value of b).

Which can also be useful when solving

|u| = a is the same as u = ±a and vice versa

Which is often the key to solving most absolute value questions.

Example: solve |x+2|=5

Using "|u| = a is the same as u = ±a":

this: |x+2|=5

is the same as this: x+2 = ±5

Which will have two solutions:

x+2 = −5 x+2 = +5

x = −7 x = 3

Graphically

Let us graph that example:

|x+2| = 5

It is easier to graph if you have an "=0" equation, so subtract 5 from both

sides:

|x+2| − 5 = 0

And here is the plot of |x+2|−5, but just for fun let's make the graph by

shifting it around:

Start with |x| then shift it left to make

it|x+2|

then shift it down to make

it|x+2|-5

And you can see the two solutions: −7 or +3.

Absolute Value Inequalities

Mixing Absolute Values and Inequalites needs a little care!

There are 4 inequalities:

< ≤ > ≥

less than less than

or equal to greater than

greater than

or equal to

Less Than, Less Than or Equal To

With "<" and "≤" you get one interval centered on zero:

Example: Solve |x| < 3

This means the distance from x to zero must be less than 3:

Everything in between (but not including) -3 and 3

It can be rewritten as:

−3 < x < 3

And as an interval it can be written as: (−3, 3)

The same thing works for "Less Than or Equal To":

Example: Solve |x| ≤ 3

Everything in between and including -3 and 3

It can be rewritten as:

−3 ≤ x ≤ 3

And as an interval it can be written as: [−3, 3]

How about a bigger example?

Example: Solve |3x-6| ≤ 12

Rewrite it as:

−12 ≤ 3x−6 ≤ 12

Add 6:

−6 ≤ 3x ≤ 18

Lastly, multiply by (1/3). Because you are multiplying by a positive

number, the inequalities will not change:

−2 ≤ x ≤ 6

Done!

And as an interval it can be written as: [−2, 6]

Greater Than, Greater Than or Equal To

This is different ... you get two separate intervals:

Example: Solve |x| > 3

It looks like this:

Up to -3 or from 3 onwards

It can be rewritten as

x < −3 or x > 3

As an interval it can be written as: (−∞, −3) U (3, +∞)

Careful! Do not write it as

−3 > x > 3

"x" cannot be less than -3 and greater than 3 at the same time

It is really:

x < −3 or x > 3

"x" is less than −3 or greater than 3

The same thing works for "Greater Than or Equal To":

Example: Solve |x| ≥ 3

Can be rewritten as

x ≤ −3 or x ≥ 3

As an interval it can be written as: (−∞, −3] U [3, +∞)

Limits (An Introduction)

Approaching

Sometimes you can't work something out directly ... but you can see what it

should be as you get closer and closer!

Let's use this function as an example:

(x2-1)/(x-1)

And let's work it out for x=1:

(12-1)/(1-1) = (1-1)/(1-1) = 0/0

Now 0/0 is a difficulty! We don't really know the value of 0/0, so we need

another way of answering this.

So instead of trying to work it out for x=1 let's try approaching it closer and

closer:

x (x2-1)/(x-1)

0.5 1.50000

0.9 1.90000

0.99 1.99000

0.999 1.99900

0.9999 1.99990

0.99999 1.99999

... ...

Now we can see that as x gets close to 1, then (x2-1)/(x-1) gets close to 2

We are now faced with an interesting situation:

When x=1 we don't know the answer (it is indeterminate)

But we can see that it is going to be 2

We want to give the answer "2" but can't, so instead mathematicians say

exactly what is going on by using the special word "limit"

The limit of (x2-1)/(x-1) as x approaches 1 is 2

And it is written in symbols as:

Right Hand Limit

x (x2-1)/(x-1)

1.5 2.50000

1.1 2.10000

1.01 2.01000

1.001 2.00100

1.0001 2.00010

1.00001 2.00001

Left Hand Limit

x (x2-1)/(x-1)

0.5 1.50000

0.9 1.90000

0.99 1.99000

0.999 1.99900

0.9999 1.99990

0.99999 1.99999

... ...

Continuous Functions

Example: f(x) = (x2-1)/(x-1) for all Real Numbers

The function is undefined when x=1:

(x2-1)/(x-1) = (12-1)/(1-1) = 0/0

So it is not a continuous function

Let us change the domain:

Example: g(x) = (x2-1)/(x-1) over the interval x<1

Almost the same function, but now it is over an interval that

does not include x=1.

So now it is a continuous function (does not include the "hole")

Introduction to Differentiation

It is all about calculating the Derivatives!

Introduction to Derivatives

It is all about slope!

Slope =

Change in Y

Change in X

You can find an average slope between two points.

We will use the slope formula:

Slope = Change in Y

= Δy

Change in X Δx

To find the derivative of a function y = f(x)

And follow these steps:

· Fill in this slope formula: Δy

=

f(x+Δx) -

f(x)

Δx Δx

· Simplify it as best you can,

· Then make Δx shrink

towards zero.

Example: the function f(x) = x2

We know f(x) = x2, and can calculate f(x+Δx) :

Start with: f(x+Δx) = (x+Δx)2

Expand (x + Δx)2: f(x+Δx) = x2 + 2x Δx + (Δx)2

Fill in the slope formula:

f(x+Δx) - f(x)

=

x2 + 2x Δx + (Δx)2 - x2

Δx Δx

Simplify (x2 and -x2 cancel): =

2x Δx + (Δx)2

Δx

Simplify more (divide through by Δx): = 2x + Δx

And then as Δx heads towards 0 we get: = 2x

Result: the derivative of x2 is 2x

Derivative Rules

Common Functions Function Derivative

Constant c 0

x 1

Square x2 2x

Square Root √x (½)x-½

Exponential ex ex

ax ax(ln a)

Logarithms ln(x) 1/x

loga(x) 1 / (x ln(a))

Trigonometry (x is in radians) sin(x) cos(x)

cos(x) -sin(x)

tan(x) sec2(x)

sin-1(x) 1/√(1-x2)

tan-1(x) 1/(1+x2)

Rules Function Derivative

Multiplication by constant cf cf’

Power Rule xn nxn-1

Sum Rule f + g f’ + g’

Difference Rule f - g f’ - g’

Product Rule fg f g’ + f’ g

Quotient Rule f/g (f’ g - g’ f )/g2

Reciprocal Rule 1/f -f’/f2

Chain Rule

(as "Composition of Functions") f º g (f’ º g) × g’

Chain Rule (in a different form) f(g(x)) f’(g(x))g’(x)

Power Rule

Example: What is x3 ?

The question is asking "what is the derivative of x3?"

We can use the Power Rule, where n=3:

xn = nxn-1

x3 = 3x3-1 = 3x2

Example: What is (1/x) ?

1/x is also x-1

We can use the Power Rule, where n=-1:

xn = nxn-1

x-1 = -1x-1-1 = -x-2

Multiplication by constant

Example: What is 5x3 ?

the derivative of cf = cf’

the derivative of 5f = 5f’

We know (from the Power Rule):

x3 = 3x3-1 = 3x2

So:

5x3 = 5 x3 = 5 × 3x2 = 15x2

Sum Rule

Example: What is the derivative of x2+x3 ?

The Sum Rule says:

the derivative of f + g = f’ + g’

So we can work out each derivative separately and then add them.

Using the Power Rule:

x2 = 2x

x3 = 3x2

And so:

the derivative of x2 + x3 = 2x + 3x2

Difference Rule

It doesn't have to be x, we can differentiate with respect to, for example, v:

Example: What is (v3-v4) ?

The Difference Rule says

the derivative of f - g = f’ - g’

So we can work out each derivative separately and then subtract them.

Using the Power Rule:

v3 = 3v2

v4 = 4v3

And so:

the derivative of v3 - v4 = 3v2 - 4v3

Sum, Difference, Constant Multiplication And Power Rules

Example: What is (5z2 + z3 - 7z4) ?

Using the Power Rule:

z2 = 2z

z3 = 3z2

z4 = 4z3

And so:

(5z2 + z3 - 7z4) = 5 × 2z + 3z2 - 7 × 4z3 = 10z + 3z2 - 28z3

Reciprocal Rule

Example: What is (1/x) ?

The Reciprocal Rule says:

the derivative of 1/f = -f’/f2

With f(x)= x, we know that f’(x) = 1

So:

the derivative of 1/x = -1/x2

Which is the same result we got above using the Power Rule.

Chain Rule

Example: What is (5x-2)3 ?

The Chain Rule says:

the derivative of f(g(x)) = f’(g(x))g’(x)

(5x-2)3 is made up of g3 and 5x-2:

f(g) = g3

g(x) = 5x-2

The individual derivatives are:

f'(g) = 3g2 (by the Power Rule)

g'(x) = 5

So:

(5x-2)3 = 3g(x)2 × 5 = 15(5x-2)2

Transcendental Function

Value is never ending cannot be calculated in a finite number of

steps

Example: What is sin(x2) ?

sin(x2) is made up of sin() and x2:

f(g) = sin(g)

g(x) = x2

The Chain Rule says:

the derivative of f(g(x)) = f'(g(x))g'(x)

The individual derivatives are:

f'(g) = cos(g)

g'(x) = 2x

So:

sin(x2) = cos(g(x)) × 2x = 2x cos(x2)