Solving the equation $f(x) = f^{-1}(x)$, where $f(x) = x^{4}-5x^{2}+x+4$ for $x\leq-1.6$

Question

Here's another question from an IB Math HL past paper -

Consider the function -

$$f(x) = x^{4}-5x^{2}+x+4, \;\;\ x\leq-1.6$$

It is given that $f^{-1}(x)$ exists.

Solve the equation $\; f(x) = f^{-1}(x)$

I have been working on this problem for almost an hour and can't figure out how to do this problem.

Any help would be highly appreciated.

NOTE: This is from a non-calculator paper.

Answer 1

On the interval $x \leq -1.6$ the function is one-one so the inverse exists. The function and it's inverse are symmetric about the line $y=x$. So for $f(x)=f^{-1}(x)$, they should both intersect at the line $y=x$. Said differently, we need to find the intersection of the function $y=f(x)$ and the line $y=x$. Thus, \begin{align*} x^{4}-5x^{2}+x+4 & = x\\ x^4-5x^2+4&=0\\ (x^2-4)(x^2-1)&=0. \end{align*} This gives $x=\pm 1, \pm 2$ as the possible solutions.But in the interval speciifed, we only have $\boxed{x=-2}$.

Answer 2

Hint: Since $f$ is invertible in the given domain, $f(x)=f^{-1}(x) \Leftrightarrow f(f(x))= f(f^{-1}(x))=x$. Therefore, $x$ a solution of $f(x)= f^{-1}(x)$ if and only if it is a solution of $f(f(x))=x$.