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January 23rd, 2016, 02:55 AM   #1
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Prove that f(x)=sin(x)+9x/π is invertible

Prove that $\displaystyle f(x)=\sin(x)+\frac{9x}{\pi }$ is invertible and find the derivative of $\displaystyle f^{-1}$ at the point 2.

Well, in order to prove the function is invertible, i have done as following:

$\displaystyle f'(x)=\cos(x)+\displaystyle\frac{9}{\pi}$
and since $\displaystyle \displaystyle\frac{9}{\pi}>1$ so the slope of the function becomes positive no matter which value the $\displaystyle x$ has, which means that the function is a strictly increasing function and therefore the $\displaystyle f$ is invertible.

So what next?

Last edited by skipjack; January 23rd, 2016 at 12:09 PM.
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January 23rd, 2016, 09:07 AM   #2
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Quote:
... at the point 2.
with the above statement, I assume you mean when $f(x)=2$ ...

$f(x) = \sin{x} + \dfrac{9x}{\pi}$

for ease of notation in typing this out, let $g(x) = f^{-1}(x)$

If $g$ and $f$ are inverses, then $g[f(x)] = x$, and ...

$g'[f(x)] \cdot f'(x) = 1$

$f(x) = 2 \implies x = \dfrac{\pi}{6}$

$g'\bigg[f\left(\dfrac{\pi}{6}\right)\bigg] = g'(2) = f^{-1}(2) = \dfrac{1}{f'\left(\dfrac{\pi}{6}\right)}$
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January 23rd, 2016, 10:39 AM   #3
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Thank you so much, with your help, I have thought as this:

$\displaystyle f(a)=b $
$\displaystyle f^{-1}(b)=a$
$\displaystyle (f^{-1})'(b)=\frac{1}{f'(a)}$
$\displaystyle b=2$, $\displaystyle a=?$

$\displaystyle f(a)=2$
$\displaystyle \sin(a)+\frac{9a}{\pi }=2$
$\displaystyle a=\frac{\pi }{6}$

$\displaystyle f'(\frac{\pi }{6})=\cos(\frac{\pi }{6})+\frac{9}{\pi }=\frac{\sqrt{3}}{2}+\frac{9}{\pi }$
$\displaystyle (f^{-1})'(2)=\frac{1}{\frac{\sqrt{3}}{2}+\frac{9}{\pi }}=\frac{2\pi }{\pi \sqrt{3}+18}$

I think this is the right answer.

Last edited by skipjack; January 23rd, 2016 at 12:37 PM.
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