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So I have a question given $f(x) = \int_1^{x^3} \sqrt{16 + t^6}dt$. The question asks to find $f^{-1'}(0)$. So I know $f^{-1'}(f(x)) = \frac{1}{f'(x)}$, so I have to solve $f(x) = 0$ or $\int_1^{x^3} \sqrt{16 + t^6}dt = 0$ first. I'm pretty sure there is a way to find the answer without having to carry out the integration because I don't have the necessary tools yet to integrate something like that, but I'm not sure how to do it.

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Let $G(t)$ be an antiderivative of $\sqrt{16+t^6}$ that you don't want to try to find. Neither do I, there is almost certainly no antiderivative that is expressible in terms of elementary functions. Then $$f(x)=G(x^3)-G(1).$$ Differentiate, using the Chain Rule. We get $$f'(x)=3x^2G'(x^3)=3x^2\sqrt{16+x^{18}}.$$ Continue, using information about how the derivative of an inverse function is connected to the derivative of the function.

Remark: We have used the Fundamental Theorem of Calculus, without explicitly mentioning it.

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+1 Thanks, I think that should be a 16 instead of a 1 underneath the square root. –  hesson Dec 16 '12 at 0:47
    
@hesson: Thank you, I am typo-prone. Do look over things, but tend not to notice things that don't matter, like constants. –  André Nicolas Dec 16 '12 at 0:49
    
Sorry, but I am still not sure how to use the derivative to solve $f(x) = 0$ which is needed to solve the question. –  hesson Dec 16 '12 at 0:57
    
But the integral is from $1$ to $x^3$, so wouldn't the solution just be $x = 1$ for $f(x) = 0$? –  hesson Dec 16 '12 at 1:13
    
Yes, I didn't look back at the equation, assumed we were starting at $0$. Indeed $f(1)=0$. –  André Nicolas Dec 16 '12 at 1:16

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