Question

Let $f:R\to R,g:R\to R$ two functions such that $f\left(x\right)=2x-3,g\left(x\right)=x^3+5$. The function ${\left(fog\right)}^{-1}\left(x\right)$ is equal to:

• A. ${\left(\frac{x+7}{2}\right)}^{\frac{1}{3}}$
• B. ${\left(x-\frac{7}{2}\right)}^{\frac{1}{3}}$
• C. ${\left(\frac{x-2}{7}\right)}^{\frac{1}{3}}$
• D. ${\left(\frac{x-7}{2}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is D

${\left(\frac{x-7}{2}\right)}^{\frac{1}{3}}$

Explanation for the correct option:

Determine the value of ${\left(fog\right)}^{-1}\left(x\right)$

We have, $f\left(x\right)=2x-3,g\left(x\right)=x^3+5$

$$\begin{gathered} fog=f\left(g\right(x\left)\right) \\ =2g\left(x\right)-3 \\ =2(x^3+5)-3 \\ =2x^3+10-3 \\ =2x^3+7 \end{gathered}$$

Now, for calculating ${\left(fog\right)}^{-1}\left(x\right)$, we will make $x$ as the subject and find the equation.

$$\begin{gathered} Let{\left(fog\right)}^{-1}\left(x\right)=y=2x^3+7 \\ \Rightarrow y=2x^3+7 \\ \Rightarrow y-7=2x^3 \\ \Rightarrow x={\left[\frac{y-7}{2}\right]}^{\frac{1}{3}} \\ \therefore {\left(fog\right)}^{-1}\left(x\right)={\left[\frac{x-7}{2}\right]}^{\frac{1}{3}} \end{gathered}$$

Hence, the correct answer is option (D).