Let $y=f(x) = -\int_1^{2/x} 2t^22^2dt$. $(t>0)$
Find the angle that $f^{-1}(x)$ make with $x$-axis.
In the book where I find this question, it was solved with that:
$f(2) = 0; f^{-1}(0)=2 \Rightarrow (f^{-1}){'}(0) = 1/f'(2)$
$f'(x) = \frac{2}{x^2}\frac{8}{x^2}2^\frac{2}{x} \therefore f'(2) = 2 \therefore (f^{-1}){'}(0) = 1/2 \therefore \theta=\tan^{-1} (1/2) $.
My problem is that I think it is not the angle with x-axis. the angle with x must be where $f^{-1}(x) = 0$ where it meets (intersects) $x$-axis. the point given in question is where it meets with $y$-axis. Am I right? If I am right, How to answer the question?