Question

1. Write the following absolute value function as a piecewise function.
f(x)= |3x+12|

Answer 1

Answer:

$|\text{3x+12}| = \begin{cases}3\text{x}+12 \ \ \ \ \ \text{ if } \text{x} \ge -4\\-3\text{x}-12\ \ \ \ \text{if } \text{x} < -4\end{cases}$

=======================================================

Explanation:

If y = |x|, then either y = x or y = -x depending on the value of x.

• If x = 0 or larger, then we go for y = x
• If x < 0, then we'll go for y = -x

We can write this as a piecewise function like this

$y = |\text{x}| = \begin{cases}\text{x} \ \text{if } \text{ x } \ge 0\\-\text{x} \ \text{if } \text{ x } < 0\end{cases}$

or like this

$|\text{x}| = \begin{cases}\text{x} \ \text{if } \text{ x } \ge 0\\-\text{x} \ \text{if } \text{ x } < 0\end{cases}$

We only involve one curly brace.

-----------

Let's replace each copy of x with (3x+12)

So that previous piecewise function updates to

$|\text{3x+12}| = \begin{cases}3\text{x}+12 \ \ \ \ \ \text{ if } 3\text{x}+12 \ge 0\\-(3\text{x}+12) \ \ \text{if } 3\text{x}+12 < 0\end{cases}$

The -(3x+12) portion distributes to -3x-12

Solving the inequality $3\text{x}+12 \ge 0$ leads to $\text{x} \ge -4$

Solving the other inequality $3\text{x}+12 < 0$ leads to $\text{x} < -4$

For each inequality, the inequality sign doesn't flip. This is because we haven't multiplied or divided both sides by a negative number.

-----------

Therefore, the piecewise function

$|\text{3x+12}| = \begin{cases}3\text{x}+12 \ \ \ \ \ \text{ if } 3\text{x}+12 \ge 0\\-(3\text{x}+12) \ \ \text{if } 3\text{x}+12 < 0\end{cases}$

is equivalent to

$|\text{3x+12}| = \begin{cases}3\text{x}+12 \ \ \ \ \ \text{ if } \text{x} \ge -4\\-3\text{x}-12\ \ \ \ \text{if } \text{x} < -4\end{cases}$

which is the final answer.

-----------

You can graph y = |3x+12| to confirm.

Desmos is a useful tool for this sort of thing. Add the linear graphs of y = 3x+12 and y = -3x-12; notice how they overlap over the V shape. This confirms that pieces of these lines form the V shape.