Problem

$$g(x)=\frac{3x^3+8x^2-3x-6}{x(x+3)}\equiv Ax+B+\frac{C}{x}+\frac{D}{x+3}$$

(a) Find the values of the constants $A$, $B$, $C$ and $D$.

A curve has equation

$$y=g(x)\qquad x>0$$

Using the answer to part (a),

(b) find $g'(x)$.

(c) Hence, explain why $g'(x)>3$ for all values of $x$ in the domain of $g$.