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Question Number 13099 by Joel577 last updated on 14/May/17

$$\mathrm{If}f(x+5)=g(2x-1)$$$$\mathrm{Find}2f^{-1}\left(x\right)$$$$$$$$\left(\mathrm{A}\right)g^{-1}\left(x\right)+11\left(\mathrm{D}\right)g^{-1}\left(x/2\right)+6$$$$\left(\mathrm{B}\right)g^{-1}\left(x\right)+9\left(\mathrm{E}\right)g^{-1}\left(2x\right)+6$$$$\left(\mathrm{C}\right)g^{-1}\left(x\right)+6$$

Answered by 433 last updated on 14/May/17

$$h\left(x\right)=x+5\mathrm{\&}k\left(x\right)=2x-1$$$$foh=gok$$$$f={gokoh}^{-1}$$$$f^{-1}={hok}^{-1}{g}^{-1}$$$$2f^{-1}=2\left({hok}^{-1}{g}^{-1}\right)$$$$k^{-1}\left(x\right)=\frac{x+1}{2}$$$$2f^{-1}\left(x\right)=2h(k^{-1}\left(g^{-1}\right)\left(x\right)$$$$=2h\left(\frac{g^{-1}\left(x\right)+1}{2}\right)$$$$=2(\frac{g^{-1}\left(x\right)+1}{2}+5)=g^{-1}\left(x\right)+1+10$$$$=g^{-1}\left(x\right)+11$$

Commented by Joel577 last updated on 15/May/17

$$thankyouverymuch$$

Answered by mrW1 last updated on 14/May/17

$$f(x+5)=g(2x-1)=u$$$$\Rightarrow x+5=f^{-1}\left(u\right)...\left(i\right)$$$$\Rightarrow 2x-1=g^{-1}\left(u\right)...\left(ii\right)$$$$2\times \left(i\right)-\left(ii\right):$$$$\Rightarrow 11=2f^{-1}\left(u\right)-g^{-1}\left(u\right)$$$$\Rightarrow 2f^{-1}\left(u\right)=g^{-1}\left(u\right)+11$$$$\Rightarrow 2f^{-1}\left(x\right)=g^{-1}\left(x\right)+11$$$$$$$$\Rightarrow Answer\left(A\right)$$

Commented by RasheedSindhi last updated on 14/May/17

$$\mathcal{E}\mathrm{asy}\mathrm{n}\mathrm{simple}!$$

Commented by Joel577 last updated on 15/May/17

$$thankyouverymuch$$

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