Problem : Let $(f \circ g)(x) = x^2 +2x -1$. Find $f(x)$ if $g(x) = x+2$

My Attempted Solution

$$f(g(x)) = x^2 +2x-1$$ $$f(x+2) = x^2 +2x -1$$

But that is as far as I got. The problem I'm having is that I can't seem to find a way to algebraically relate $g(x)$ to $f(g(x))$ and solve for the unknown $f(x)$.

As a generalized extension to the above problem, furthermore if I have a composition of two aribtrary functions $f,g : \mathbb{R} \to \mathbb{R}$ with $g$, known and $f$ being an unknown function, what is the general algebraic method to solve for $f(x)$ if $f(g(x))$ is given. i.e.

$$\text{If}\ \ f(g(x)) = \alpha \ \ \text{and} \ \ g(x) = \zeta \ \ $$ $$\text{How do you algebraically solve for } f(x)$$

  • $\begingroup$ Hint: $x= (x+2)-2$ $\endgroup$ – John Joy Jun 4 '16 at 2:15

Hint: $x^2 + 2x - 1 = (x+2)^2 - 2(x+2) - 1$

  • $\begingroup$ is there a more general way to solve this sort of problem? You've manipulated $x^2 + 2x-1$ into a form that is meaningful to $f(x+2)$, but it seems like you've done it in a sort of trial-and-error way (a heuristic way). If you hadn't spotted that possible manipulation /transformation (as I didn't), how would you have attempted to solve the problem? (Note: This is aimed at the second part of my question in the OP) $\endgroup$ – Perturbative Jun 3 '16 at 18:37
  • $\begingroup$ @Perturbative Here the inside function $g$ is just a "shift". You can use a substitution as suggested by the other answerer; this will always work. More generally, if you have $h(x) = f(g(x))$ and you know $g(x)$ and $h(x)$, then, given that $g$ is a "nice" function (a bijection $\mathbb{R} \to \mathbb{R}$ so we have no domain issues), $f(x) = h(g^{-1}(x))$. As a worked example, take $f \circ g \ (x) = \exp(x)$ and $g(x) = x^3$. Then $f(x)= \exp(x^{\frac 13})$ $\endgroup$ – MathematicsStudent1122 Jun 3 '16 at 18:45

Replace $x \rightarrow x -2$ in $f(x+2) = x^2 +2x -1$

  • $\begingroup$ More fundamentally: If $(f\circ g)(x)=h(x)$, then $f(x)=(f\circ g\circ g^{-1})(x)=(h\circ g^{-1})(x)$. $\endgroup$ – Semiclassical Jun 3 '16 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.