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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
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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 ...

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

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

$g'\bigg[f\left(\dfrac{\pi}{6}\right)\bigg] = g'(2) = f^{-1}(2) = \dfrac{1}{f'\left(\dfrac{\pi}{6}\right)}$ 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. Tags 9x or, 9x or π, fxsinx, invertible, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post chaarleey Calculus 6 June 4th, 2015 08:03 AM Magnesium Linear Algebra 2 December 11th, 2013 02:09 AM problem Linear Algebra 3 August 31st, 2011 05:30 AM tinynerdi Linear Algebra 0 February 20th, 2010 05:58 PM rbaptista Linear Algebra 1 November 23rd, 2008 01:58 PM

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