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This function passes the horizontal line test so there does exist an inverse for it. The problem I am running into is simplifying the equation when I interchange y and x. $$ y=x^3+4x-1 $$ I have got it into the form of the inverse pre-requisite: $$x=y^3+4y-1$$. The most I have progressed with the problem is moving the "1" to the other side of the equation and somewhat simplifying the cubic expression: $$ x+1 = y(y^2-4) $$ This is the initial part of the problem. The second part asks to find $ f^{-1}(4) $. For this, I substituted 4 into the previous equation to get $$ 5 = y(y^2-4) $$.

Edit: I wish to find $$f^{-1}(x)$$ and $$f^{-1}(4)$$.

Obaid Jaffery
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Well if you need to find $f^{-1}(x)$ then there's only one way to find the inverse of a function like $y^3+4y-1=x$ is by treating $x$ like a constant, $$y^3+4y-(x+1)=0$$ and use the formula for finding the roots of a depressed cubic polynomial, $$y=r+s-\dfrac 43$$ where $r$ and $s$ are the roots of the quadratic, $$u^2 +(16+3(x+1))u - (x+1)^2=0$$

After all the hardwork you will get the required $f^{-1}(x)$ to the polynomial. But if your motive is to find $f^{-1}(4)$ then you don't need to overwork yourself.

You can substitute $y=4$ in $y=x^3+4x-1$ and find the solution for $x$ which you will get $x=1$ as the only real solution. Thus you have $f^{-1}(4)=1$

Learningstill
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If you follow the staps given here, using (why ?) the hyperbolic solution, you should need almost one minute to obtain $$x=\frac{4}{\sqrt{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3\sqrt{3}}{16}(y+1)\right)\right)$$ which is a bit nicer than what Wolfram Alpha produces.

Claude Leibovici
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