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November 8th, 2018, 04:15 PM   #11
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 Originally Posted by Chemist116  Sorry I wasn't aware of that. But from looking at the messages it looks that $x=\frac{11}{12}$. But does it means that there are other solutions as well?. I missed this part. Which would be the other solution?.
Compare posts 2 and 3. November 8th, 2018, 05:06 PM   #12
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 Originally Posted by Chemist116 $\phi^{-1}\left(\phi\left( 2x+1\right)\right)=\phi^{-1}\left(6x-10\right)$ $2x+1= \frac{y+10}{6}$
That isn't a correct way to obtain the inverse. If you look carefully at what you did, you may be able to see why.

Once you have the correct inverse, namely $\phi^{-1}(x) = (x + 13)/3$,
applying it to the second equation gives $f(x) - 3 = ((3x - 4) + 13)/3 = x + 3$, so $f(x) = x + 6$. Tags embedded, functions, method, problem, shown, work 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 LIPTON Physics 1 June 28th, 2012 09:33 AM Kimmysmiles0 Algebra 2 April 29th, 2012 08:38 PM Arley Algebra 1 March 22nd, 2012 07:51 PM Recipe Algebra 3 March 17th, 2010 06:10 PM Arley Calculus 0 December 31st, 1969 04:00 PM

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