I came across the following problem:
Find all $x$ values such that the tangent to the function $f(x) = \frac{1}{x^2+1} + (1-2x)^{1/3}$ where $x \ge 0$ at that $x$ value is perpendicular to the tangent of its inverse function $f^{-1}(x)$ at that $x$ value.
I know that a formula for the derivative of the inverse function is
$(f^{-1})^{'}(x) = \frac{1}{f'(f^{-1}(x))}$ and we want this expression for $(f^{-1})^{'}(x)$ to be the negative reciprocal of $f'(x)$ for the tangent of the function to be perpendicular to the tangent of its inverse at $x$. This gives the equation $f'(x) = -f'(f^{-1}(x))$. However, it seems the only way to solve this equation for $x$ is to find $f^{-1}(x)$ explicitly and I can't seem to find the inverse of $f$ because it is so complicated. I tried the standard swapping $y$ and $x$ and solving for $y$ trick. I even tried Wolfram Alpha but it didn't seem to find an answer in terms of elementary functions.
Is there a way to solve this problem without explicitly finding the inverse or, if not, how would one go about finding the inverse of $f(x) = \frac{1}{x^2+1} + (1-2x)^{1/3}$ where $x \ge 0$ ?
Best Answer
There is no such $x$. The function $f$ is strictly decreasing and any tangent line to its graph has negative slope, with one exception: the tangent line at $\left(\frac12,\frac45\right)$ is a vertical line. So, $f^{-1}$ is strictly decreasing too and any tangent line to its graph has negative slope, again with one exception: the tangent line at $\left(\frac45,\frac12\right)$ is a horizontal line. So, these two lines are the only example of a tangent line to the graph of $f$ and a tangent line to the graph of $f^{-1}$ which are perpendicular. But $\frac45\ne\frac12$, and so the problem has no solution.