Find the composition f(g(x)) when four functions are given

Question

Determine $f \circ g$ for the following functions: $$ f(x) = \begin{cases} -x, & x < 0\\ x+1, & x \ge 0 \end{cases} \quad \text{and} \quad g(x) = \begin{cases} x^2, & x \le 2\\ x+2, & x > 2 \end{cases} $$

I need some help with this exercise. I have no issues with finding the composition $f(g(x))$. But I don't have an idea how to do it when four functions are given.

Answer 1

HINT

Well, if $x > 2$ then $g(x) = x+2 > 4$ so $f(g(x)) = (x+2)+1$.

Otherwise, if $x \le 2$, then $g(x) = x^2 \le 4$ and $g(x) \ge 0$ since a square of a number is always positive. Can you finish this?

Answer 2

The method:

First determine for which values of $x$ the function $f(x)\leq 2$ and for which $f(x)< 2$.

For this you should:

1. Assume that $x<0$ and then determine when $-x\leq 2$ and when $-x>2$
2. Assume that $x\geq 0$ and then determine when $x+1\leq 2$ and when $x+1>2$

Once you do that you can compose with $g$ more easily. The part for which $f(x)\leq 2$ is being composed with $x^2$ (i.e. you square $f(x)$ if it is less or equal $2$.) and the part for which $f(x)>2$ is being composed with $x+2$ (i.e. you add $2$ to $f(x)$ if it is greater than $2$).

At the end of the day you will have a lot of cases. But this can't be avoided.

Answer 3

What you can do is break this problem apart and create a new composition, with intervals $x < 0, 0 \leq x < 2, x \geq 2$. On each of these intervals, your functions are normal, and as you said you can compute the normal composition. And yes, it will yield a giant case function :).