If $f(x)=\int_{3}^x \sqrt{1+t^3} dt$ Find $(f^{-1})'(0)$

Question

Consider the integral $$f(x)=\int_{3}^x \sqrt{1+t^3} dt$$

Using Theorem 7 from my textbook which is $$\frac{1}{f'(f^{-1}(a))}$$

$f'(t)= \frac{2t^3}{2\sqrt{1+t^3}}$

$f^{-1}(t)=\sqrt[3]{t^2-1}$

I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?

Answer 1

HINTS:

From the Fundamental Theorem of Calculus, we have

$$f'(x)=\sqrt{1+x^3}$$

Moreover, note that

$$f(3)=0\implies f^{-1}(0)=3$$

Finally, the theorem from your text book reveals that

$$\frac{df^{-1}(x)}{dx}=\frac1{\sqrt{1+(f^{-1}(x))^3}}$$