Inverse Function of a Closed Interval

Question

This is an example from the book Topology 2nd edition by Munkers. Example 4 page 20.

Please see the example here

How is it that $f^{-1}(2) = -1$? Shouldn't $f^{-1}(2) = 0$? Since $f(x) = 3x^2 + 2 \implies f^{-1}(x) = \sqrt{\frac{x-2}{3}}$?

Answer 1

Note that .. $$f^{-1}(x)=\pm \sqrt{\frac{x-2}{3}}$$

And in the picture it is given range of inverse function which is true at $x=5$ . You are right that inverse function at $x=2$ is $0$ . $$f^{-1}(2)=0$$ But that doesn't necessarily gives you the minimum value at this value. You have to check the Maxima and minima which are obtained at $x=5$

Hence $$f^{-1}([2,5])=[-1,1] $$ is correct in the given picture.