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

The inverse of the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$ is

Let $R$ denote the set of all real numbers. Let $f: R \rightarrow R$ and $g: R \rightarrow (0, 4)$ be functions defined by $f(x) = \log_e(x^2 + 2x + 4)$ and $g(x) = \frac{4}{1 + e^{-2x}}$. Define the composite function $h(x) = (f \circ g^{-1})(x)$,where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $h(x)$ at $x = 2$ is:

$f: R \rightarrow R, f(x) = 3x + 2$ and $g: R \rightarrow R, g(x) = 6x + 5$. Find the value of $(g \circ f^{-1})(10)$.

Let $f : A \to B$ be a function defined as $f(x) = \frac{x - 1}{x - 2}$,where $A = R - \{2\}$ and $B = R - \{1\}$. Then $f$ is

If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = 3x - 4$,then ${f^{ - 1}}: \mathbb{R} \to \mathbb{R}$ is