Compute ($f^{-1})'(3)$ if $f(x) = x^2 + \tan(\pi x/2)$ for $-1 < x < 1$

Question

I'm not sure if this is a typo from my Calculus 2 homework, but I know that for the equation $x^2 + \tan(\pi x/2) = 0$, $x = 0$. I can't figure out how to calculate $x$ when $x^2 + \tan(\pi x/2)= 3$. The only way I can think of answering this question is by graphing it on a calculator and finding the intersection. Is there another way to do this? Thanks!

Answer 1

Since we're restricted to the range $[-1, 1]$, which due to multiplying by $\pi/2$ is one period of the function, we can assume the inverse is a single function with a single value at $x=3$.

Therefore, we can take the derivative of the given function at the point where f(x) = 3, then take the multiplicative inverse of that result.

As a simpler example, take the function $f(x) = x^3$. Its value at $f(x) = 8$ is $x=2$. Therefore at $f(x) = 8$,

$$ f(x)' = 3x^2 = 12 $$

Therefore $(f^{-1})'(8) = 12^{-1} = \frac{1}{12} $. We can do this because $f^{-1}(x)$ is a reflection across the line $y = x $ on a typical Cartesian graph.

(Edited: mixed up a 2 and an 8)

Answer 2

Beside purely numerical methods, such as Newton's, if you know the values of trigonometric functions at particulat angles, you should notice that the solution is close to $x=\frac 3 4$ since $$\left(\frac{3}{4}\right)^2+\tan \left(\frac{3\pi }{8}\right)=\frac 9 {16}+\cot \left(\frac{\pi }{8}\right)\approx 2.97671$$

So, to have an approximate solution, expand $f(x)$ as a Taylor series around $x=\frac 34$. Since we know the exact values for $\frac \pi 8$ (see here), the expansion will be $$\left(\frac{25}{16}+\sqrt{2}\right)+\left(\frac{3}{2}+\left(2+\sqrt{2}\right ) \pi \right) \left(x-\frac{3}{4}\right)+O\left(\left(x-\frac{3}{4}\right)^{2}\right)$$ and, ignoring the higher order tems, an estimate is $$x=\frac 3 4+\frac{23-16 \sqrt{2}}{24+16 \left(2+\sqrt{2}\right) \pi }\approx 0.751905$$ while the exact solution is $0.751892$.

Using the $[1,1]$ (the simplest) Padé approximant around $x=\frac 34$, we should have $$x=\frac 3 4+\frac{69-48 \sqrt{2}+\left(28-18 \sqrt{2}\right) \pi }{118-32 \sqrt{2}+96 \left(2+\sqrt{2}\right) \pi +\left(188+133 \sqrt{2}\right) \pi ^2}$$ which is $0.751892$ ! Bingo !!