If $g$ is the inverse of a function $f$ and $f'(x) = \frac{1}{1 + x^5}$,then $g'(x)$ is equal to:

$f: R \rightarrow R$,$f(x) = 4x + 3$ is defined,then $f^{-1}(x) =$ . . . . . . .

Let $f: R \rightarrow R, g: R \rightarrow R$ and $h: R \rightarrow R$ be differentiable functions such that $f(x)=x^3+3x+2, g(f(x))=x$ and $h(g(g(x)))=x$ for all $x \in R$. Then

Let $A = \{1, 2, 3\}$ and $B = \{1, 3, 5\}$. $A$ relation $R: A \to B$ is defined by $R = \{(1, 3), (1, 5), (2, 1)\}$. Then ${R^{-1}}$ is defined by:

If $f = \{(1,2), (2,3), (3,1)\}$,it is easy to see that $f$ is one-one and onto. Find the inverse function $f^{-1}$.

If $f(x) = x^2 + 1$,then $f^{-1}(17)$ and $f^{-1}(-3)$ will be