Find Value of $g'(0)$ if $g(x)$ is inverse of $f(x)$, where $f(x)=\int_{2}^{x}\frac{1}{\sqrt{(1+t^4)}}dt$.

Question

Find Value of $g'(0)$ if $g(x)$ is inverse of $f(x)$ where $$f(x)=\int_{2}^{x}\frac{1}{\sqrt{(1+t^4)}}dt.$$

I had tried following things

finding $f(x)$ by integration but failed
is $g(x)$ is inverse of $f(x)$ then $$fog = x$$ So $$f'(g(x)) = \frac{1}{g'(x)}$$ and differentiate $f(x)$ by applying Newton - Leibniz and get $$g'(x) = {\sqrt{1+{g(x)}^{4}}}$$ And after putting $$ x = 0$$ I get $$g'(0) = {\sqrt{1+{g(0)}^{4}}}$$

But still I need to find $$g(0)$$ And I have no clue how to find that

@zahbaz sorry it's a mistake I am correcting it . – Mohit Sep 16 '18 at 21:34 — Mohit, Sep 16 '18 at 21:34

zahbaz · Accepted Answer · 2018-09-16T21:45:02.987

4

If $g= f^{-1}$, then when $f(x)=0$, $g(0)=x$. We can solve for it then using

$$0=\int_{2}^{g(0)}\frac{1}{{(1+t^4)}^{1/2}}dt$$

No need to integrate. What must $g(0)$ be?

$g(0)=2$. The integrand is positive, so the interval of integration must vanish.

edited Sep 16 '18 at 21:45

answered Sep 16 '18 at 21:39

zahbaz

Thanks buddy for helping me – Mohit Sep 16 '18 at 21:41
Happy to help :) – zahbaz Sep 16 '18 at 21:42
Would it be when the function being integrated has $t=g(0)$ and this is equal to when $t=2$? – Henry Lee Sep 16 '18 at 21:44
1

@HenryLee Yes, though for clarity, I'd use a variable other than $t$, since that's the integration variable. See spoiler box in edit for solution. – zahbaz Sep 16 '18 at 21:45

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