If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^3}$,then $g^{\prime}(x)$ is

Let $g(x)$ be the inverse of an invertible function $f(x)$ which is differentiable at $x = c$. Then $g'(f(c))$ equals:

Let $S = \{1, 2, 3\}$. Determine whether the function $f: S \rightarrow S$ defined as below has an inverse. Find $f^{-1}$,if it exists: $f = \{(1, 1), (2, 2), (3, 3)\}$.

Let $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.
Statement-$1$: $S = \{x : f(x) = f^{-1}(x)\} = \{1, 2\}$.
Statement-$2$: $f$ is a bijection and $f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$.

The number of integral values of $a$ for which the function $f: R \to R, f(x) = 2x^3 - 3(a + 2)x^2 + 12ax - 7$ where $a \in [-4, 6]$ is invertible,is

If $f(x) = \frac{2x - 1}{x + 5}, x \neq -5$ then $f^{-1}(x)$ is equal to