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If $f(x)=\frac{2}{3} x+\frac{3}{2}, x \in R, \quad$ [2010-II]
then what is $f^{-1}(x)$ equal to?
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then what is $f^{-1}(x)$ equal to?
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Verified Answer
The correct answer is:
$\frac{3}{2} x-\frac{9}{4}$
Let $f(x)=\frac{2}{3} x+\frac{3}{2}=y($ say $)=\frac{4 x+9}{6}=y$
$\Rightarrow \quad 4 x+9=6 y$
$\Rightarrow \quad x=\frac{6 y-9}{4}$
$\quad x=f^{-1}(y)$
$\Rightarrow \quad f^{-1}(x)=\frac{6 x-9}{4}=\frac{3 x}{2}-\frac{9}{4}$
$\Rightarrow \quad 4 x+9=6 y$
$\Rightarrow \quad x=\frac{6 y-9}{4}$
$\quad x=f^{-1}(y)$
$\Rightarrow \quad f^{-1}(x)=\frac{6 x-9}{4}=\frac{3 x}{2}-\frac{9}{4}$
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