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

Consider $f: R_{+} \rightarrow [4, \infty)$ given by $f(x) = x^{2} + 4$. Show that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}(y) = \sqrt{y - 4}$,where $R_{+}$ is the set of all non-negative real numbers.

Let $g(x)$ be the inverse of the function $f(x)$ and $f'(x) = \frac{1}{1 + x^3}$. Then $g'(x)$ is equal to

Let $f(x)=(x+1)^2-1$ for $x \geq -1$.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.

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

Which of the following functions is invertible?