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Question Number 165949 by mathlove last updated on 10/Feb/22

$$f\left(x\right)=\frac{5x-2}{a}\wedge f^{-1}\left(x\right)=\frac{x+b}{5}$$$$fainda\times b=?$$

Answered by qaz last updated on 10/Feb/22

$${\left(\frac{5\mathrm{x}-2}{\mathrm{a}}\right)}^{\prime }\cdot {\left(\frac{\mathrm{x}+\mathrm{b}}{5}\right)}^{\prime }=\frac{1}{\mathrm{a}}=1$$$$\Rightarrow \mathrm{a}=1$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=5\mathrm{x}-2$$$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{2+\mathrm{x}}{5}=\frac{\mathrm{x}+\mathrm{b}}{5}$$$$\Rightarrow \mathrm{b}=2$$$$\Rightarrow \mathrm{a}\times \mathrm{b}=2$$

Answered by cortano1 last updated on 10/Feb/22

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{5\mathrm{x}-2}{\mathrm{a}}\iff {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{-\mathrm{ax}-2}{-5}=\frac{\mathrm{ax}+2}{5}$$$$\mathrm{so}\{\begin{array}{l}\mathrm{a}=1\\ \mathrm{b}=2\end{array}\Rightarrow \mathrm{a}\times \mathrm{b}=2$$

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