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If a function $f: R \rightarrow R$ is defined by $f(x)=\frac{4 x}{5}+3$, then $f^{-1}(x)=$
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The correct answer is:
$\frac{5(x-3)}{4}$
(A)
Let $f(x)=\frac{4 x}{5}+3=y$
$\therefore 4 x=5 y-15 \Rightarrow x=\frac{5 y-15}{4}$
$\therefore f^{-1}(y)=\frac{5 y-15}{4} \Rightarrow f^{-1}(x)=\frac{5 x-15}{4}=\frac{5(x-3)}{4}$
Let $f(x)=\frac{4 x}{5}+3=y$
$\therefore 4 x=5 y-15 \Rightarrow x=\frac{5 y-15}{4}$
$\therefore f^{-1}(y)=\frac{5 y-15}{4} \Rightarrow f^{-1}(x)=\frac{5 x-15}{4}=\frac{5(x-3)}{4}$
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