Determine $g(x)$ such that $(f\circ g)(x) = x^2 - 4x + 5$

Question

Consider $f(x) = x^2 + 2x + 5$ . Now find $g(x)$ such that $(f\circ g)(x) = x^2 - 4x + 5$ .

Note : I know that it will be easy if we suppose $g(x) = ax+b$ and then solve it but I'm curious about a general method . I mean solving $(g(x))^2 + 2g(x) + 5 = x^2 - 4x + 5$ . And also for all cases , is finding $g(x)$ possible if we have $(f\circ g)(x)$ and $f(x)$ ?

Answer 1

hint

Assume that $$f (g (x))=h (x) $$

for $x\in A .$

If $f $ is a bijection from $g (A) $ to $h (A) $ then

$$g (x)=f^{-1}(h (x)) $$

In your case,

$$f (x)=x^2+2x+5$$ $$f^{-1}(x)=-1\pm \sqrt{x-4} $$

Answer 2

From $(g(x))^2 + 2g(x) + 5 = x^2 - 4x + 5$,

\begin{align} [g(x)+1]^2&=x^2-4x+1\\ g(x)&=-1\pm\sqrt{x^2-4x+1} \end{align}

For the second question, it depends on $f$. If $f$ is bijective, then $g(x)=f^{-1}(f(g(x)))$.