If $f: R \rightarrow R$ is given by $f(x)=7 x+8$ and $f^{-1}(12)=\frac{k}{7}$, then the value of $k$ is

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Aicharya · Accepted Answer

We have $f(x)=7 x+8=y$...(let) $	herefore \mathrm{x}=\frac{\mathrm{y}-8}{7} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{\mathrm{y}-8}{7}$ $	herefore \mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}-8}{7} \Rightarrow \mathrm{f}^{-1}(12)=\frac{12-8}{7}=\frac{4}{7} \Rightarrow \mathrm{k}=4$

If $f: R \rightarrow R$ is given by $f(x)=7 x+8$ and $f^{-1}(12)=\frac{k}{7}$, then the value of $k$ is

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