Consider: $$g(x)=x^{1/3}$$ $$f(g(x))=2x+3$$
Find $f(x)$
It should be fairly obvious that a solution is $f(x)=2x^3+3$
But is this solution unique? And what method should I use to solve such an equation if I don't "see" the solution?
3 Answers
Suppose we have an equation $f(g(x)) =h(x) $ where $g$ and $h$ are known, and $g$ has an inverse. Defining $y=g(x)$, we have $f(y) =h(g^{-1} (y))$, which is the solution. If $g$ does not have a unique inverse (that is, the inverse of $g(x) $ depends on x), then the solution for $f$ will not be unique.
It suffices to invert $g$:
$$y=x^{1/3}\iff x=y^3$$
and there is no "other" inverse.
Then
$$f(x)=f(g(g^{-1}(x)))=f(g(x^3))=2x^3+3.$$
Yes, the solution is unique. You have two functions, $g:\mathbb{R}\rightarrow \mathbb{R}$ and $f:\mathbb{R}\rightarrow \mathbb{R}$.
For every $x$, $g(x)=\sqrt[3]{x}$. You have the cube root function, and it is a bijection.
For every $x$, $f(g(x))=2x+3$. Then $f(\sqrt[3]{x})=2x+3$ for every $x$. Since cube root is a bijection, you can conclude that $f(x)=2x^3+3$.
