Finding two functions $f(x)$ and $g(x)$

Question

I am not sure how to approach this question. It asks to find $f(x)$ and $g(x)$ such that $h(x)=f(g(x))$, for each function:

a) $$h(x)=\sqrt{x^2 + 6}$$

b)$$h(x)=\frac{1}{x^3}-7x+2$$

If someone could help me, that would be really appreciated.

Answer 1

Let's look at a few different ways to do the first one, then maybe you'll see how to do the second:

1. Let $f(x) = \sqrt{x^2 + 6}$ and $g(x) = x$. Then $$f(g(x)) = f(x) = \sqrt{(x)^2+6} = h(x)$$
2. That one sorta feels like it's cheating, though, doesn't it? Let's find a not so trivial way to do it. How about $f(x) = \sqrt{x+6}$ and $g(x) = x^2$. Then $$f(g(x)) = f(x^2) = \sqrt{(x^2)+6} = h(x)$$
3. Or how about this one? Let $f(x) = \sqrt{x}$ and $g(x)=x^2+6$. Then $$f(g(x)) = f(x^2+6) = \sqrt{(x^2+6)} = h(x)$$ That one makes both functions really simple -- wouldn't you agree?

As you can see there are multiple ways of doing these types of compositions. It'd be a good exercise if now you try to find an $f$ and $g$ different than any of my examples above. So give it a try. :)

Answer 2

Well, you can always take $g(x)=x$, and $f(x) = h(x)$. Then, you have $f(g(x)) = h(g(x)) = x$ and you are done.