Question

Let $f:R\to R$ be a function defined by $f\left(x\right)=x^3+x^2+x-1.$ If $g$ is the inverse of $f,$ then $g^{\prime }\left(2\right)$ is

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

Answer 1

The correct option is C $\frac{1}{6}$

We have, $g\left(x\right)=f^{-1}\left(x\right)\Rightarrow f\left(g\right(x\left)\right)=x$
Differentiating both sides w.r.t. $x,$ we get
$f^{\prime }\left(g\right(x\left)\right)\times g^{\prime }\left(x\right)=1$
$f^{\prime }\left(x\right)=3x^2+2x+1$

Let $g\left(2\right)=K$
We know $f\left(g\right(2\left)\right)=2$
$\Rightarrow f\left(K\right)=2$
$\Rightarrow K^3+K^2+K-1=2$
$\Rightarrow K^3+K^2+K-3=0$
$\therefore K=1\Rightarrow g\left(2\right)=1.$
$g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\right(x\left)\right)}$
$g^{\prime }\left(2\right)=\frac{1}{f^{\prime }\left(g\right(2\left)\right)}=\frac{1}{f^{\prime }\left(1\right)}=\frac{1}{6}(\because f^{\prime }(1)=6)$