Find $g \circ f$ with $f$ and $g$ given.

Question

Suppose $f(x)=\begin{cases}x+1 & x \ge 1 \\ x+2 & x<1\end{cases}$ and $g(x)=\begin{cases}x^2 & x\ge 1 \\ x^3 & x<1 \end{cases} $

What is $g(f(x))$?

I'm confused.

Answer 1

Hint: Before you can compute $g(f(x))$, you will need to determine which values of $x$ produce $f(x)\leq 1$ and which values of $x$ produce $f(x)>1$.

That has to be done so you will know which part of $g$ to use when applying it to $f(x)$.

Examine each piece in the definition of $f$ to do this.

To help you get started, consider the first piece of $f$, the part where $f(x)=x+1$ when $x\geq 1$. Among those values of $x$, you want to find which ones produce $f(x)\geq 1$ (if any) and which ones produce $f(x)<1$ (if any). Use the definition of $f$ on this part of the domain to see that you are really asking "for what values of $x\geq 1$ is $\overbrace{x+1}^{\textrm{value of }f(x)}\underbrace{\leq 1}_{\textrm{domain of }g}$?" and "for what values of $x\geq 1$ is $\overbrace{x+1}^{\textrm{value of }f(x)}\underbrace{>1}_{\textrm{domain of }g}$?".

Then repeat this procedure for the other part of $f$.

Answer 2

So if you are given two functions $f(x) = x + 3$ and $g(x) = 9x^{2}$, for example, here is how to find $f(g(x))$ and $g(f(x))$.

You look at the outside function, and replace the $x$'s with a $\Box$. So for example, if we want to find $f(g(x))$, the outside function is $f(x) = x + 3$. So I will rewrite this as $f(\Box) = \Box + 3$. Now, everywhere I see a $\Box$, I need to write what $g(x)$ is. but $g(x) = 9x^{2}$, so that means we get $f(g(x)) = 9x^{2} + 3$.

Similarly, to find $g(f(x))$, the outside function is $g(x) = 9x^{2}$. I need to replace all of the $x$'s with a $\Box$, and so I get $g(\Box) = 9 \Box^{2}$. Now I need to write what $f(x)$, the inside function, is in each of the boxes. I get $g(f(x)) = 9 (x + 3)^{2}$.

Finally, you do the same thing with the problem you have. Except you have a piecewise function. So you do the same thing, but you look at both functions for $x \geq 1$ at first, and do that piece, and then do the $x < 1$ piece of both functions second. Your answer will also be a piecewise function.