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

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

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

The condition for a function $y = f(x)$ to be invertible is that it must be:

If $\alpha$ is the minimum value for which the inverse of $f(x)=x^2+3x-3$ exists in $[\alpha, \infty)$ and $g$ is the inverse of $f$, then find the value of $\frac{dg}{dx}$ at $x=\alpha+\frac{5}{2}$.

If the functions $f$ and $g$ are defined by $f(x) = 3x - 4$ and $g(x) = 2 + 3x$ for $x \in R$,then $g^{-1}(f^{-1}(5))$ is equal to