Question:
If $f: R \rightarrow R, g: R \rightarrow R$ are defined by $f(x)=5 x-3, g(x)=x^{2}+3$, then $\left(g \circ f^{-1}\right)(3)=$
Solution:
Given: $f(x)=5 x-3$ and $g(x)=x^{2}+3$
$f(x)=5 x-3$
$\Rightarrow y=5 x-3$
$\Rightarrow 5 x=y+3$
$\Rightarrow x=\frac{y+3}{5}$
Thus, $f^{-1}(y)=\frac{y+3}{5}$.
Now,
$g o f^{-1}(3)=g\left(f^{-1}(3)\right)$
$=g\left(\frac{3+3}{5}\right)$
$=g\left(\frac{6}{5}\right)$
$=\left(\frac{6}{5}\right)^{2}+3$
$=\frac{36}{25}+3$
$=\frac{36+75}{25}$
$=\frac{111}{25}$
Hence, gof $^{-1}(3)=\frac{111}{25}$.
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