Derivatives and Inverses!

$f(x) = ax^{b} \implies f'(x) = abx^{b - 1}$ and $f^{-1}(x) = \dfrac{1}{a^{(\dfrac{1}{b})}} x^{\dfrac{1}{b}}$

For $x^{b - 1} = x^{\dfrac{1}{b}} \implies \boxed{b - 1 = \dfrac{1}{b}} \implies b^2 - b - 1 = 0$ and

$b > 0 \implies b = \dfrac{1 + \sqrt{5}}{2} = \phi$

For $ab = \dfrac{1}{a^{(\dfrac{1}{b})}} = \dfrac{1}{a^{b - 1}} \implies a^{b} b = 1 \implies$ $a = (\dfrac{1}{b})^{\dfrac{1}{b}}$

Checking:

$f'(x) = (\dfrac{1}{b})^{\dfrac{1}{b}} b x^{b - 1} = b^{(\dfrac{b - 1}{b})} x^{b - 1}$

and

$f^{-1}(x) = (b^{\dfrac{1}{b}})^{(\dfrac{1}{b})} x^{b - 1} = b^{(\dfrac{b - 1}{b})} x^{b - 1} = f'(x)$ .

$\therefore f(x) = (\phi - 1)^{\phi - 1} x^{\phi}$ and $f^{-1}(x) = \phi^{(\dfrac{\phi - 1}{\phi})} x^{\phi - 1}$

The area $A = \displaystyle\int_{0}^{\phi} f^{-1}(x) - f(x) \:\ dx = \dfrac{\phi^{(\dfrac{\phi - 1}{\phi})}}{\phi} x^{\phi} - \dfrac{(\phi - 1)^{\phi - 1} x^{\phi + 1}}{\phi^2}|_{0}^{\phi} =$

$\phi^{(\dfrac{\phi^2 - 1}{\phi})} - \dfrac{\phi^{\phi + 1}}{\phi^{(\dfrac{2\phi + 1}{\phi})}}$

$= \phi^{(\dfrac{\phi + 1 - 1}{\phi})} - \phi^{(\dfrac{\phi^2 - \phi - 1}{\phi})} = \boxed{\phi - 1}$

and from above $f^{-1}(\phi) = \phi^{(\dfrac{\phi - 1}{\phi})} (\phi)^{\phi - 1} =$ $(\phi)^{(\dfrac{\phi^2 - 1}{\phi})} = (\phi)^{(\dfrac{\phi + 1 - 1}{\phi})} = \boxed{\phi}$

$\implies f^{-1}(\phi) - A = \phi - (\phi - 1) = \boxed{1}$ .

Exact same question as https://brilliant.org/problems/inverse-and-derivative/ .