Question

If $f\left(\right.x\left.\right)=x^{3}+3x+4$ and $g$ is the inverse function of $f$ , then the value of $\frac{d}{d x}\left(\frac{g \left(\right. x \left.\right)}{g \left(\right. g \left(\right. x \left.\right) \left.\right)}\right)$ at $x=4$ is

• A. $\frac{- 1}{3}$
• B. $\frac{- 1}{2}$
• C. $3$
• D. $6$

Answer 1

Answer: (A)

Let $D=\frac{d}{d x}\left(\frac{g(x)}{g(g(x))}\right)$ at $x=4$
$ \left.=\frac{g(g(x)) g^{\prime}(x)-g(x) \cdot g^{\prime}(g(x)) \cdot g^{\prime}(x)}{(g(g(x))}\right]_{x=4} $
Now, $f(x)=x^{3}+3 x+4$
$ \Rightarrow f^{\prime}(x)=3 x^{2}+3>0 $
Clearly, $f(x)$ is an increasing function.
Now, $f(0)=4 \Rightarrow f^{-1}(4)=g(4)=0$
Also, $g(f(x))=x$
$ \begin{array}{l} \therefore g^{\prime}(f(x)) f^{\prime}(x)=1 \\ \Rightarrow g^{\prime}(f(0)) f^{\prime}(0)=1 \\ \Rightarrow g^{\prime}(4)=\frac{1}{f^{\prime}(0)}=\frac{1}{3} \end{array} $
$f(-1)=0 \Rightarrow f^{-1}(0)=g(0)=-11$
$\therefore D=\frac{g(g(4)) g^{\prime}(4)-g(4) g^{\prime}(g(4)) g^{\prime}(4)}{(g(g(4))}$
$=\frac{g(0) \cdot \frac{1}{f^{\prime}(0)}-0}{(g(0)}=\frac{(-1) \times \frac{1}{3}}{1}=\frac{-1}{3}$