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

Let the function $f$ be defined by $f(x) = \frac{2x + 1}{1 - 3x}$,then $f^{-1}(x)$ is

Let $f: R - \{3\} \rightarrow R - \{1\}$ be defined by $f(x) = \frac{x-2}{x-3}$. Let $g: R \rightarrow R$ be given as $g(x) = 2x - 3$. Then,the sum of all the values of $x$ for which $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$ is equal to ...... .

Consider $f: R_{+} \rightarrow [-5, \infty)$ given by $f(x) = 9x^{2} + 6x - 5$. Show that $f$ is invertible with $f^{-1}(y) = \frac{\sqrt{y+6}-1}{3}$.

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

If $f: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+1}{5x-3}$,then which of the following is true?