Let $f(x)=(x+1)^2-1, x \geqslant-1$,then the set $\{x : f(x)=f^{-1}(x)\}$ is

$f: (-\infty, 0] \rightarrow [0, \infty)$ is defined as $f(x) = x^2$. The domain and range of its inverse are

Suppose $f(x) = (x + 1)^2$ for $x \ge -1$. If $g(x)$ is the function whose graph is the reflection of the graph of $f(x)$ with respect to the line $y = x$,then $g(x)$ equals

Let $f:(2, 3) \to (0, 1)$ be defined by $f(x) = x - [x]$. Then ${f^{ - 1}}(x)$ equals:

If $f: N \rightarrow N$ and $f(x) = x + 3$,then $f^{-1}(x) =$ . . . . . . .

Let $g$ be the inverse function of $f$ and $f'(x) = \frac{x^{10}}{1 + x^2}$. If $g(2) = a$,then $g'(2)$ is equal to: