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Question Number 124444 by benjo_mathlover last updated on 03/Dec/20

$$Iff\left(\frac{2}{x}\right)=\frac{3}{2+4x}and{f}^{-1}\left(p\right)=2$$$$thenp=?$$

Answered by bemath last updated on 03/Dec/20

$$\iff f^{-1}\left(p\right)=2thenf\left(2\right)=p$$$$weget\{\begin{array}{l}\frac{2}{x}=2;x=1\\ p=\frac{3}{2+4x}=\frac{3}{2+4\left(1\right)}=\frac{1}{2}\end{array}$$

Answered by mathmax by abdo last updated on 03/Dec/20

$${\mathrm{f}}^{-1}\left(\mathrm{p}\right)=2\Rightarrow \mathrm{p}=\mathrm{f}\left(2\right)\mathrm{but}\mathrm{f}\left(\frac{2}{\mathrm{x}}\right)=\frac{3}{2+4\mathrm{x}}\mathrm{let}\mathrm{take}\mathrm{x}=1\Rightarrow$$$$\mathrm{f}\left(2\right)=\frac{3}{6}=\frac{1}{2}\Rightarrow \mathrm{p}=\frac{1}{2}$$

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