Suppose that $g(x) = 1 + \sqrt{x}$ and $f(g(x)) = 3 + 2\sqrt{x} + x$,then $f(x)$ is

If $f(x) = 3x + 10$ and $g(x) = x^2 - 1$,then $(fog)^{-1}$ is equal to

If $f(x)=e^x$ and $g(x)=\ln(x)$ for all $x \in [1, \infty)$,then $f \circ g$ is . . . . . .

Let $(g \circ f)(x) = \sin x$ and $(f \circ g)(x) = (\sin \sqrt{x})^2$. Then,

If $f(x) = \frac{3x+4}{5x-7}, x \neq \frac{7}{5}$ and $g(x) = \frac{7x+4}{5x-3}, x \neq \frac{3}{5}$,then $(g \circ f)(3) = $

If $f(x) = \begin{cases} \sin x, & x \ne n\pi, n \in \mathbb{Z} \\ 2, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \ne 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x))$ is