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If $f(x)=3 x-5$, then $f^{-1}(x)$
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Verified Answer
The correct answer is:
Is given by $\frac{x+5}{3}$
Let $f(x)=y \Rightarrow x=f^{-1}(y)$.
Hence $f(x)=y=3 x-5 \Rightarrow x=\frac{y+5}{3} \Rightarrow f^{-1}(y)=x=\frac{y+5}{3}$
$\therefore f^{-1}(x)=\frac{x+5}{3}$
Also $f$ is one-one and onto, so $f^{-1}$ exists and is given by $f^{-1}(x)=\frac{x+5}{3}$
Hence $f(x)=y=3 x-5 \Rightarrow x=\frac{y+5}{3} \Rightarrow f^{-1}(y)=x=\frac{y+5}{3}$
$\therefore f^{-1}(x)=\frac{x+5}{3}$
Also $f$ is one-one and onto, so $f^{-1}$ exists and is given by $f^{-1}(x)=\frac{x+5}{3}$
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