If $y = f(x) = \frac{x + 2}{x - 1}$,then $x = $

If $f : R \to R$ is defined by $f(x) = x^2 + 1$,then $f^{-1}(17)$ and $f^{-1}(-3)$ are

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 $x=\alpha+\frac{5}{2}$.

Consider the function $y = \log_{a}(x + \sqrt{x^{2} + 1})$ where $a > 0$ and $a \neq 1$. The inverse of the function:

If $f(x) = (x + 1)^2$ for $x \ge 1$, and $g(x)$ is a function whose graph is the reflection of the graph of $f(x)$ in the line $y = x$, then $g(x)$ is:

If ${e^x} = y + \sqrt {1 + {y^2}} $,then $y =$