Let $f$ be a function such that $f(x + y) = f(x) + f(y)$ for all $x$ and $y$,and $f(x) = (2x^2 + 3x)g(x)$ for all $x$,where $g(x)$ is continuous and $g(0) = 3$. Then $f'(x)$ is equal to:

Let $f: R^{+} \rightarrow R^{+}$ be a function satisfying $f(x) - x = \lambda$ (constant),$\forall x \in R^{+}$ and $f(x f(y)) = f(x y) + x, \forall x, y \in R^{+}$. Then $\lim _{x \rightarrow 0} \frac{(f(x))^{\frac{1}{3}} - 1}{(f(x))^{\frac{1}{2}} - 1} =$

Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x, y \in R$,and $g: R \rightarrow(0, \infty)$ be a function such that $g(x+y)=g(x) g(y)$ for all $x, y \in R$. If $f\left(\frac{-3}{5}\right)=12$ and $g\left(\frac{-1}{3}\right)=2$,then the value of $\left(f\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)$ is.

$f: R \rightarrow R$ is defined by $f(x+y)=f(x)+12y, \forall x, y \in R$. If $f(1)=6$,then $\sum_{r=1}^n f(r)=$

Let $f(x)$ be defined for all $x > 0$ and be continuous. Let $f(x)$ satisfy $f\left( \frac{x}{y} \right) = f(x) - f(y)$ for all $x, y > 0$ and $f(e) = 1$. Then:

$A$ function $f: R \rightarrow R$ satisfies $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)+f(0)}{3}$ for all $x, y \in R$. If the function $f$ is differentiable at $x=0$, then $f$ is: