Introduction to The Modulus Function

Since $f(x) = x^2 - 2x = x(x-2)$ , theb $f(x) < 0$ when $0 < x < 2$ and we have:

$|f(x)| = \begin{cases} x(2-x) = 2x - x^2 & \text{if }0 \le x < 2 \\ x(x-2) = x^2 - 2x & \text{elsewhere} \end{cases}$

Note that

$g(|x|) = |x|^2 - 4|x| + 4 = \begin{cases} x^2 - 4x + 4 & \text{if } x \ge 0 \\ x^2 + 4x + 4 & \text{if } x < 0 \end{cases}$

Therefore

$|f(x) = g(|x|) = \begin{cases} x^2-2x = x^2 + 4x + 4 \implies x = - \frac 23 & \text{if }x < 0 \\ 2x- x^2 = x^2 - 4x + 4 \implies x = 1 & \text{if }0 \le x < 2 \\ x^2-2x = x^2 - 4x + 4 \implies x = 2 & \text{if }x \ge 2 \end{cases}$

There are $\boxed 3$ solutions.

Read about absolute values here .

Here's a more visual approach to the problem.

First, let's sketch the graphs of $f(x), |f(x)|, g(x)$ and $g(|x|)$ .

You can sketch $f(x)$ and $g(x)$ by solving $f(x)=0$ and $g(x)=0$ to find the $x$ intercepts , and letting $x=0$ to find the vertical intercept .

Then, to sketch $|f(x)|$ , reflect any parts of the graph of $f(x)$ that are below the $x$ axis in the $x$ axis .

Finally, to sketch $g(|x|)$ , sketch $g(x)$ for $x=0$ and $x>0$ , and then reflect this in the $y$ axis .

This process is shown below:

If we superimpose the graphs of $|f(x)|$ and $g(|x|)$ , we can see where they intersect :

Bonus : find the coordinates of A , B and C .