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January 23rd, 2016, 02:55 AM  #1 
Newbie Joined: Jan 2016 From: sweden Posts: 11 Thanks: 0  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. 
January 23rd, 2016, 09:07 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,940 Thanks: 1545  Quote:
$f(x) = \sin{x} + \dfrac{9x}{\pi}$ 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 ... $g'[f(x)] \cdot f'(x) = 1$ $f(x) = 2 \implies x = \dfrac{\pi}{6}$ $g'\bigg[f\left(\dfrac{\pi}{6}\right)\bigg] = g'(2) = f^{1}(2) = \dfrac{1}{f'\left(\dfrac{\pi}{6}\right)}$  
January 23rd, 2016, 10:39 AM  #3 
Newbie Joined: Jan 2016 From: sweden Posts: 11 Thanks: 0 
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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9x or, 9x or π, fxsinx, invertible, prove 
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