Question

If $f\left(x\right)=2x^3+7x-5andg\left(x\right)=f^{-1}\left(x\right)$ then g' (4) is equal to

• A. $\frac{1}{13}$
• B. $\frac{1}{103}$
• C. $\frac{1}{4}$
• D. non existent

Answer 1

The correct option is A $\frac{1}{13}$

Given, $f\left(x\right)=2x^3+7x-5$

$\therefore f^{\prime }\left(x\right)=6x^2+7$

$g\left(x\right)=f^{-1}\left(x\right)\Rightarrow g\left\{f\right(x\left)\right\}=x$
Differentiating both sides, w.r.t $x$
$g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}=\frac{1}{6x^2+7}$
when $x=1,f\left(x\right)=4$
$\therefore g^{\prime }\left(4\right)=\frac{1}{13}$