Question:
If $f(x)=3 x+10$ and $g(x)=x^{2}-1$, then $(f \circ g)^{-1}$ is equal to
Solution:
$f o g(x)=f(g(x))$
$=f\left(x^{2}-1\right)$
$=3\left(x^{2}-1\right)+10$
$=3 x^{2}-3+10$
$=3 x^{2}+7$
Thus, $\operatorname{fog}(x)=3 x^{2}+7$
$\Rightarrow y=3 x^{2}+7$
$\Rightarrow 3 x^{2}=y-7$
$\Rightarrow x^{2}=\frac{y-7}{3}$
$\Rightarrow x=\pm \sqrt{\frac{y-7}{3}}$
$f o g^{-1}(x)=\pm \sqrt{\frac{x-7}{3}}$
Hence, $(\text { fog })^{-1}$ is equal to $\pm \sqrt{\frac{x-7}{3}}$.
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