If $f(x)$ and $g(x)$ are two functions such that $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$,then $f'(1)$ is equal to:

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

Consider functions $f: A \rightarrow B$ and $g: B \rightarrow C$ $(A, B, C \subseteq \mathbb{R})$ such that $(g \circ f)^{-1}$ exists. Then:

If $f(x)=3x+6$,$g(x)=4x+k$ and $f \circ g(x)=g \circ f(x)$,then $k =$

If $f(x) = \frac{\alpha x}{x + 1}$,$x \ne -1$,for what value of $\alpha$ is $f(f(x)) = x$?

Let $f, g$ and $h$ be functions from $R$ to $R$. Show that:
$\begin{cases} (f+g)oh = foh + goh \\ (f \cdot g)oh = (foh) \cdot (goh) \end{cases}$