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Let $f: R \rightarrow R, g: R \rightarrow R$ be two functions such that
$f(x)=2 x-3, g(x)=x^{3}+5$. The function $(f \circ g)^{-1}(x)$ is equal to
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$f(x)=2 x-3, g(x)=x^{3}+5$. The function $(f \circ g)^{-1}(x)$ is equal to
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Verified Answer
The correct answer is:
$\quad\left(\frac{x-7}{2}\right)^{1 / 3}$
We have, $f(x)=2 x-3, g(x)=x^{3}+5$
$(f \circ g) x=f\left(x^{3}+5\right)=2\left(x^{3}+5\right)-3=2 x^{3}+7$
Let $y=(f \circ g) x=2 x^{3}+7$
$$
\Rightarrow \mathrm{x}=\left(\frac{\mathrm{y}-7}{2}\right)^{1 / 3} \Rightarrow(\text { fog })^{-1} \mathrm{x}=\left(\frac{\mathrm{x}-7}{2}\right)^{1 / 3}
$$
$(f \circ g) x=f\left(x^{3}+5\right)=2\left(x^{3}+5\right)-3=2 x^{3}+7$
Let $y=(f \circ g) x=2 x^{3}+7$
$$
\Rightarrow \mathrm{x}=\left(\frac{\mathrm{y}-7}{2}\right)^{1 / 3} \Rightarrow(\text { fog })^{-1} \mathrm{x}=\left(\frac{\mathrm{x}-7}{2}\right)^{1 / 3}
$$
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