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

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

Let $f: X \rightarrow Y$ be an invertible function. Show that $f$ has a unique inverse.
(Hint: Suppose $g_{1}$ and $g_{2}$ are two inverses of $f$. Then for all $y \in Y$,$f \circ g_{1}(y) = I_{Y}(y) = f \circ g_{2}(y)$. Use the one-one property of $f$.)

The inverse of the function $f(x) = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$ is

Let $f(x) = \frac{x^2 - x}{x^2 + 2x}$,$x \ne 0, -2$. Then $\frac{d}{dx}[f^{-1}(x)]$ (wherever it is defined) is equal to

Let $f : R \rightarrow R$ be defined by $f(x) = x^3 - 3x^2 + 3x - 2$. Then $f^{-1}(x)$ is given by: