Suppose $f(x)=(x+1)^{2}$ for $x \geq -1$. If $g(x)$ is a function whose graph is the reflection of the graph of $f(x)$ in the line $y=x$,then $g(x) = $

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

Let $f: X \rightarrow Y$ be an invertible function. Show that $f$ has a unique inverse.
(Hint: Suppose $g_{1}$ and $g_{2}$ are two inverses of $f$. Then for all $y \in Y$,$f \circ g_{1}(y) = I_{Y}(y) = f \circ g_{2}(y)$. Use the one-one property of $f$.)

Let $f : R \to R$ be defined by $f(x) = \ln(x + \sqrt{x^2 + 1})$. Then the number of solutions of $|f^{-1}(x)| = e^{-|x|}$ is

Let $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. Then $g(f(x))$ is invertible for $x \in $

If the function $f(x)=x^3+e^{\frac{x}{2}}$ and $g(x)=f^{-1}(x)$,then the value of $g^{\prime}(1)$ is