Absolute Value Function

This is the Absolute Value Function: f(x) = |x|. It is also sometimes written: abs(x).

f(x)=|x|,  the absolute Value Function can be defined as a piece wise function,

|x| = x ,if x >0 or -x if x<0  or 0 if x=0

Illustration 

Find |-5|

According to the definition |x|= -x if x<0

So |-5|=-(-5)=5

In other words |x|>0 (or equal to)

Problem Solving strategy 

1) Drawing the graph.

2) Algebraic method. 

Graphical Method 

How to draw the graph 

Shortcut

If you need to draw the graph of |y|, (y=f(x))then simply draw the graph of y=f(x)

Now in the graph of y=f(x) if anything lies in the negative y axis, just imagine a mirror in the x axis and draw mirror image of the of part lying in the negative y axis on the positive y axis. 

Illustration 

Draw the graph of y=|x³|

 Draw the graph of y=x³

Now think about the mirror 

Resulting graph

|x³|

Illustration 

Solve ||4x-3|-15|=|2x²-3x-27|

Hint draw the graph

Inequalities including absolute value function 

|x|>a , if a∈positive reals.

x∈(-∞,a)U(a,∞)

if |x|<a ,  a∈positive reals.

-a<x<a

If |x|>a , a∈negative reals.

Then it is true for all real x.

|x|<a, a∈negative reals. 

It is not true for any real x.

Problem 

Find the minimum integral value of a, such that:

|x-a|+2a < 2021,

x has 5! values. 

Problem solving strategy for inequality involving absolute value function 

|y| , y=f(x)

|y|>0

To remove the absolute value function, just square it , and science a²≥0 we can remove the abs. function.

|y|≥0

y²≥0

Problem

2|x|-|x-2|≤0

HINT

2|x|≤|x-2|

4x²≤(x-2)²

Problems from IOQM Digatal camp