If the function $f:[1, \infty) \to [1, \infty)$ is defined by $f(x) = 2^{x(x - 1)}$,then $f^{-1}(x)$ is

If $f(x) = x^2 - x + 5, x > \frac{1}{2},$ and $g(x)$ is its inverse function,then $g'(7)$ equals

Let $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ be a one-one and onto function given by $f(1) = a$,$f(2) = b$,and $f(3) = c$. Show that there exists a function $g: \{a, b, c\} \rightarrow \{1, 2, 3\}$ such that $g \circ f = I_X$ and $f \circ g = I_Y$,where $X = \{1, 2, 3\}$ and $Y = \{a, b, c\}$.

Let $f: [4, \infty) \to [1, \infty)$ be a function defined by $f(x) = 5^{x(x - 4)}$. Then $f^{-1}(x)$ is:

Let $f(x) = x^5 + 2e^{x/4}$ for all $x \in \mathbb{R}$. Consider a function $g(x)$ such that $(g \circ f)(x) = x$ for all $x \in \mathbb{R}$. Then the value of $8g'(2)$ is:

Let $x \neq 0$ and $|x| < \frac{1}{2}$. If $f(x) = 1 + 2x + 4x^2 + 8x^3 + \ldots$, then $f^{-1}(x) =$