Let $$f : R \to R$$ be defined as $$f(x) = x^3 + x - 5$$. If $$g(x)$$ is a function such that $$f(g(x)) = x, \forall x \in R$$, then $$g'(63)$$ is equal to

We are given $$f(x) = x^3 + x - 5$$ and $$f(g(x)) = x$$, so $$g$$ is the inverse function of $$f$$.

By the inverse function theorem, $$g'(y) = \frac{1}{f'(g(y))}$$.

To find $$g'(63)$$, we first need $$g(63)$$, i.e., the value $$a$$ such that $$f(a) = 63$$:

$$ a^3 + a - 5 = 63 $$

$$ a^3 + a = 68 $$

Testing $$a = 4$$: $$4^3 + 4 = 64 + 4 = 68$$ ✓

So $$g(63) = 4$$.

Now compute $$f'(x) = 3x^2 + 1$$, so $$f'(4) = 3(16) + 1 = 49$$.

Therefore:

$$ g'(63) = \frac{1}{f'(g(63))} = \frac{1}{f'(4)} = \frac{1}{49} $$

The answer is Option B: $$\frac{1}{49}$$.