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

If $f(x) = (x+1)^2 - 1$ for $x \geq -1$,then find the set $\{x \mid f(x) = f^{-1}(x)\}$.

If $f:[1, +\infty) \to [2, +\infty)$ is given by $f(x) = x + \frac{1}{x}$,then ${f^{-1}}(x)$ equals:

Let $f$ be a real-valued function defined on the interval $(-1, 1)$ such that $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt$,for all $x \in (-1, 1)$ and let $f^{-1}$ be the inverse function of $f$. Then $(f^{-1})'(2)$ is equal to

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