Let f: R → R be defined as f(x) = x ^ 3 + x - 5 . If g(x) is a...

Here, $g\left(f\left(x\right)\right) = x$ as $f\left(x\right)$ and g \left(x\right) are inverse of each other.

Now, {g}^{'}\left(f\left(x\right)\right){f}^{'}\left(x\right) = 1$</p><p>$⇒ {g}^{'}\left(f \left(x\right)\right) = \dfrac{1}{{f}^{'}\left(x\right)} \ldots \left(i\right)

Now $f\left(x\right) = 63 ⇒ {x}^{3} + x - 5 = 63$

$⇒ {x}^{3} + x - 68 = 0$

So $x = 4$ satisfies the above equation$</p><p>${g}^{'}\left(63\right) = \dfrac{1}{{f}^{'}\left(4\right)}$from \left(i\right)$

$= \dfrac{1}{3{\left(4\right)}^{2} + 1}=\dfrac{1}{49}$