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

For $x \in R$,two real-valued functions $f(x)$ and $g(x)$ are such that $g(x) = \sqrt{x} + 1$ and $(f \circ g)(x) = x + 3 - \sqrt{x}$. Then $f(0)$ is equal to:

Let $f: R \rightarrow R$ be a function defined by $f(x) = \left(2\left(1 - \frac{x^{25}}{2}\right)\left(2 + x^{25}\right)\right)^{\frac{1}{50}}$. If the function $g(x) = f(f(f(x))) + f(f(x))$,then the greatest integer less than or equal to $g(1)$ is

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

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