Question

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

Answer 1

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