Integrals and Inverse Functions

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Homework Statement


Set f(x)=\int^{2x}_{1}\sqrt{16 + t^{4}}dt.
A. Show that f has an inverse.
B. Find (f^{-1})'(0).


Homework Equations


(f^{-1})'(x)=1/(f'(f^{-1}(x)))


The Attempt at a Solution


A. f'(x)=\sqrt{16 + t^{4}} >0, so f is always increasing, hence one-to-one. By definition of inverse functions, f would have an inverse.

B. (f^{-1})'(0)=1/(f'(f^{-1}(0)))

I'm not sure where to go from here for part B. Help?
 
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I think maybe you can find the inverse of the function firstly,then subtitute x by 0 ? I am also focusing on part B
 
(Solving f(x) = 0 should be easy...)

You did forget something, though: to have an inverse, a function must be one-to-one and onto. So, for example, you would need to show that \lim_{x \rightarrow +\infty} f(x) = +\infty[/itex], and similarly for -\infty.
 
ak123456 said:
I think maybe you can find the inverse of the function firstly,then subtitute x by 0 ? I am also focusing on part B

Thanks, I think I've got it. I was overlooking some details.
 
It is not necessary to actually determine f or f-1. Since f is defined by an integral, the "fundamental theorem of calculus" gives you its derivative.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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