Question

f(x)=integral from 0 to x sqrt(1+t^4), show that f has an inverse and find (f^-1)'(c), c=f(1)

Answer 1

There is this weird identity for taking the derivative of an inverse function.\[\Large \left[f^{-1}\right]'(a)\quad=\quad \frac{1}{f'\left[f^{-1}(a)\right]}\]
And we're trying to find,\[\Large \left[f^{-1}\right]'(c)\]So maybe we want to use that. Hmm.

Answer 2

\[\Large f(1)=c \qquad\to\qquad f^{-1}(c)=1\]Yes...? +_+ Maybe? I think..

Answer 3

\[\Large \left[f^{-1}\right]'(c) \quad=\quad \frac{1}{f'\left[\color{orangered}{f^{-1}(c)}\right]}\]

Answer 4

See where we're going with this? :U

Answer 5

kind of, should we take the integral and take the inverse of that

Answer 6

\[\Large \left[f^{-1}\right]'(c) \quad=\quad \frac{1}{f'\left[\color{orangered}{f^{-1}(c)}\right]}\]Well we know what f^{-1}(c) is, so we can plug it in:\[\Large \left[f^{-1}\right]'(c) \quad=\quad \frac{1}{f'\left[\color{orangered}{1}\right]}\]

Answer 7

Now finding f'(1) is a little tricky.
Gotta remember back to our `Fundamental Theorem of Calculus, Part 1`.
\[\Large \frac{d}{dx}\int\limits_0^x f(t)\;dt \quad=\quad f(x)\]

Answer 8

\[\Large f(x)\quad=\quad \int\limits_0^x \sqrt{1+t^4}\;dt\]
Applying the FTC,\[\Large f'(x)\quad=\quad \sqrt{1+x^4}\]