Let $f: R \rightarrow R$ be defined by $f(x) = 2x + 6$,which is a bijective mapping. Then $f^{-1}(x)$ is given by:

If $f(x) = \frac{4x+3}{6x-4}, x \neq \frac{2}{3},$ show that $(f \circ f)(x) = x$ for all $x \neq \frac{2}{3}.$ What is the inverse of $f$?

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 $f(x)=\sin 2x + \cos 2x$ and $g(x)=x^2-1$. Then $g(f(x))$ is invertible in the domain:

Let $f(x) = (x + 2)^2 - 2, x \geq - 2$. Then $f^{-1}(x) =$

Let $e^{f(x)} = \ln x$. If $g(x)$ is the inverse function of $f(x)$,then $g'(x)$ is equal to: