Find $g(x)$ given $f(x)$ and the composition $(g \circ f)(x)$

Question

I've been stuck on this final math problem for ages

I'm given $$f(x) = x^2 + 1$$

and the final composition is $$(g \circ f)(x) = \frac{1}{x^2 + 4}.$$

I'm asked to find that $g(x)$ was in order to make this true, but I'm not sure how?

Answer 1

Here's my reasoning:

$$ g(f(x)) = \frac{1}{x^2 + 4} = \frac{1}{(x^2+1)+3}$$

Since $x^2 + 1 = f(x)$

$$g(f(x)) = \frac{1}{f(x)+3} \implies g(x) = \frac{1}{x+3} $$

With $g(x)$, Note: $x \ne -3$ $(x \in \mathbb{R})$

Answer 2

For general case, it is better to use inverse function concept. One of the characteristic of the inverse function is the function map to itself when compose to its inverse i.e. suppose $f(x)$ is a function and $f^{-1}(x)$ is the inverse function of $f(x)$, then $f[f^{-1}(x)]=x$.

We will propose a theorem for solving general case of finding the function of the left side of the composite function.

Theorem: If $fg(x)=h(x)$ and $g(x)$ are well defined functions, then $f(x)=h[g^{-1}(x)]$.

Given that $g(x)$ and $fg(x)$ are defined as functions. The composite of two functions with one variable will become another function with one variable hence $fg(x)=h(x)$ for some function of $h(x)$.

Hence $x$ is substituted with $g^{-1}(x)$ and yield: $$f(g[g^{-1}(x)])=h[g^{-1}(x)].$$

By applying $g[g^{-1}(x)]=x$ (inverse function property) into the previous equation, we get: $$f(x)=h[g^{-1}(x)].$$

Hence, the general case of finding $f(x)$ is proven. $\blacksquare$

Coming back to the question, to find $g(x)$ from given composite function $gf(x)$ and $f(x)$, the $f^{-1}(x)$ must be find first. If $f(x)$ is not restricted to any domain, then the theorem that have been proved cannot be use since $f(x)$ is not a bijective function. However, if $f(x)$ has stricter domain such as $x \geq 0$, then $f(x)$ has inverse function. Hence, the theorem can be applied.

Finding the inverse function of $f(x)$ which is $f^{-1}(x)=\sqrt{x-1}$ and substituted in $gf(x)$. Hence $g(x)$ is: $$\begin{aligned} g(x)&=\frac{1}{(f^{-1}(x))^2+4}\\&=\frac{1}{(\sqrt{x-1})^2+4}\\&=\frac{1}{x+3},x\neq-3. \end{aligned}$$

Answer 3

This becomes way easier if you have the knowledge of inverse functions.

Let me make this clear. You have $f(x) = x^2 + 1$ and $g(f(x)) = 1/(x^2 + 4)$. Now pause and think about the second function. The function is defined as $g(f(x))$, right. now what if there is some way that you could manipulate this function and some how change it to $g(x)$. Or think about with what other function should you multiply $f(x)$ [the function in $g(f(x))$ ] to get $x$ [so that the function would be $g(x)$]. It is $f(x)^{-1}$ right.

So let us find $f(x)$ inverse. $f(x) = x^2 +1$ then $f(x)^{-1} = \sqrt{x - 1}$ right [ just swipe the $x$ with $f(x)$ and solve for $f(x)$ that should give you this result] now go to the first function $g(f(x))$ [this function is the same as $g(x^2 + 1)$ right.]

Now find $g(\sqrt{x -1})$ (that is $g$ of the inverse function you have found) in the function which is $g( x^2 + 1)$.

Thus the result becomes $$g((\sqrt{x - 1})^2 + 1) = \frac{1}{(\sqrt{x - 1})^2 + 4}.$$ Now here comes the beauty: what is $(\sqrt{x-1})^2 + 1$? It is $x$, so the value of $g((\sqrt{x-1})^2 + 1) = g(x)$. Now $$g(x) = \frac{1}{(\sqrt{x - 1})^2 + 4} = \frac{1}{x + 3 }.$$

Answer 4

$g(f(x)) = \frac{1}{x^2 + 4}$

$g(x^2 + 1) = \frac{1}{x^2 + 4}$

How to go from $(x^2 + 1)$ to $\frac{1}{x^2 + 4}$?

1. $(x^2 + 1) + 3$ (Add $3$)
2. $\frac{1}{(x^2 + 1) + 3}$ (Take the reciprocal)

Ergo, $g(x) = \frac{1}{x + 3}$