Question

Let $f\left(x\right)=x^2-4x-3\text{,}x>2$ and $g(x)$ be the inverse of $f(x)$ . Then the value of $\frac{1}{g^{{}^{\prime }}(2)}$ , where $f(x)=2$ , is (here, $g^{{}^{\prime }}$ represents the first derivative of $g$ )

Answer 1

Answer: 6

As $g(f(x))=x$ $(\because g(x)=f^{-1}(x))$
Differentiating, we get,
$g^{{}^{\prime }}(f(x))f^{{}^{\prime }}(x)=1$
$\Rightarrow$ $g^{{}^{\prime }}(f(x))=\frac{1}{f^{{}^{\prime }}(x)}$
when $f(x)=2$
$\Rightarrow$ $x^2-4x-3=2$
$\Rightarrow$ $x=5,x=-1$
$\because x>2$
$\therefore x=5$
$g^{{}^{\prime }}(2)=\frac{1}{(2x-4{)}_{x=5}}$
$=\frac{1}{6}$