If f(x) = 3x + 10 and g(x)

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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