If $f(x) = (x+1)^2 - 1$ for $x \geq -1$,then find the set $\{x \mid f(x) = f^{-1}(x)\}$.

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

Let $f : A \to B$ be a function defined as $f(x) = \frac{x - 1}{x - 2}$,where $A = R - \{2\}$ and $B = R - \{1\}$. Then $f$ is

Which of the following functions cannot have their inverse defined? (where $[.] \to$ greatest integer function)

Let $R$ denote the set of all real numbers. Let $f: R \rightarrow R$ and $g: R \rightarrow (0, 4)$ be functions defined by $f(x) = \log_e(x^2 + 2x + 4)$ and $g(x) = \frac{4}{1 + e^{-2x}}$. Define the composite function $h(x) = (f \circ g^{-1})(x)$,where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $h(x)$ at $x = 2$ is:

If $f: R \rightarrow R$ is defined by $f(x)=2x+3$,then $f^{-1}(x)$