1. 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.

2. 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|

3. 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:
is the same as this:
|x+2|=5
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:

4. then shift it left to make
it|x+2|
Start with |x|
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

5. 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:
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
[−2, 6]

6. It can be rewritten as
x < −3
or
As an interval it can be written as:
x>3
(−∞, −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
As an interval it can be written as:
x≥3
(−∞, −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:

7. (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:

8. 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
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
...
...
Left Hand Limit
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

9. 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.

10. We will use the slope formula:
Slope =
Change in Y
Change in X
=
Δy
Δx
To find the derivative of a function y = f(x)
And follow these steps:
· Fill in this slope formula:
Δy
=
Δx
f(x+Δx) f(x)
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
x2 + 2x Δx + (Δx)2 - x2
f(x+Δx) - f(x)
Fill in the slope formula:
=
Δx
Δx
2x Δx + (Δx)2
Simplify (x2 and -x2 cancel):
=
Δx
Simplify more (divide through by Δx):
= 2x + Δx

11. And then as Δx heads towards 0 we get:
= 2x
Result: the derivative of x2 is 2x
Derivative Rules
Common Functions
Function
Derivative
c
0
x
1
Square
x2
2x
Square Root
√x
(½)x-½
Exponential
ex
ex
ax
ax(ln a)
ln(x)
1/x
loga(x)
1 / (x ln(a))
sin(x)
cos(x)
cos(x)
-sin(x)
tan(x)
sec2(x)
sin-1(x)
1/√(1-x2)
Constant
Logarithms
Trigonometry (x is in radians)

12. tan-1(x)
1/(1+x2)
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")
fºg
(f’ º g) × g’
Chain Rule (in a different form)
f(g(x))
f’(g(x))g’(x)
Rules
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

13. 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

14. 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

15. 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) is made up of g3 and 5x-2:
3


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)

16. The individual derivatives are:


f'(g) = cos(g)
g'(x) = 2x
So:
sin(x2) = cos(g(x)) × 2x = 2x cos(x2)