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If $f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ are two functions defined by $f(x)=2 x-3, \mathrm{~g}(x)=x^{3}+5$
then $(\operatorname{fog})^{-1}(x)=$
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then $(\operatorname{fog})^{-1}(x)=$
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Verified Answer
The correct answer is:
$\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
Given $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3, \quad \mathrm{~g}(\mathrm{x})=\mathrm{x}^{3}+5$
$\begin{aligned} \therefore \quad(\mathrm{fog})(\mathrm{x}) &=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right) \\ &=2\left(\mathrm{x}^{3}+5\right)-3=2 \mathrm{x}^{3}+7 \end{aligned}$
$\begin{aligned} & \text { Let } y=2 x^{3}+7 \Rightarrow \frac{y-7}{2}=x^{3} \\ \therefore &\left(\frac{y-7}{2}\right)^{\frac{1}{3}}=x \Rightarrow f^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\ &(f \circ g)^{-1}(x)=\left(\frac{x-7}{2}\right)^{\frac{1}{3}} \end{aligned}$
$\begin{aligned} \therefore \quad(\mathrm{fog})(\mathrm{x}) &=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right) \\ &=2\left(\mathrm{x}^{3}+5\right)-3=2 \mathrm{x}^{3}+7 \end{aligned}$
$\begin{aligned} & \text { Let } y=2 x^{3}+7 \Rightarrow \frac{y-7}{2}=x^{3} \\ \therefore &\left(\frac{y-7}{2}\right)^{\frac{1}{3}}=x \Rightarrow f^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\ &(f \circ g)^{-1}(x)=\left(\frac{x-7}{2}\right)^{\frac{1}{3}} \end{aligned}$
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