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If $\mathrm{f}: \mathrm{IR} \rightarrow \mathrm{IR}, \mathrm{g}: \mathrm{IR} \rightarrow \mathrm{IR}$ be two functions given by
$\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{3}+5$, then $(\mathrm{fog})^{-1}(\mathrm{x})$ is equal to
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$\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{3}+5$, then $(\mathrm{fog})^{-1}(\mathrm{x})$ is equal to
Solution:
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Verified Answer
The correct answer is:
$\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
$\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3$
$\mathrm{~g}(\mathrm{x})=\mathrm{x}^{3}+5$
$\operatorname{fog}(\mathrm{x})=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right)$
$=2\left(\mathrm{x}^{3}+5\right)-3$
$\operatorname{fog}(\mathrm{x})=2 \mathrm{x}^{3}+7=\mathrm{y}(\mathrm{say})$
$\Rightarrow 2 \mathrm{x}^{3}=\mathrm{y}-7$
$\Rightarrow \mathrm{x}=\left[\frac{\mathrm{y}-7}{2}\right]^{\frac{1}{3}}$
$\Rightarrow \operatorname{fog}^{-1}(\mathrm{y})=\left[\frac{\mathrm{y}-7}{2}\right]^{\frac{1}{3}}$
$\Rightarrow \mathrm{fog}^{-1}(\mathrm{x})=\left[\frac{\mathrm{x}-7}{2}\right]^{\frac{1}{3}}$
$\mathrm{~g}(\mathrm{x})=\mathrm{x}^{3}+5$
$\operatorname{fog}(\mathrm{x})=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right)$
$=2\left(\mathrm{x}^{3}+5\right)-3$
$\operatorname{fog}(\mathrm{x})=2 \mathrm{x}^{3}+7=\mathrm{y}(\mathrm{say})$
$\Rightarrow 2 \mathrm{x}^{3}=\mathrm{y}-7$
$\Rightarrow \mathrm{x}=\left[\frac{\mathrm{y}-7}{2}\right]^{\frac{1}{3}}$
$\Rightarrow \operatorname{fog}^{-1}(\mathrm{y})=\left[\frac{\mathrm{y}-7}{2}\right]^{\frac{1}{3}}$
$\Rightarrow \mathrm{fog}^{-1}(\mathrm{x})=\left[\frac{\mathrm{x}-7}{2}\right]^{\frac{1}{3}}$
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