Long Question 3

Question 3:
Solution:
(a)
$\begin{array}{l}\text{Given gradient function of a curve }\frac{dy}{dx}=6x^2-12x\\ \text{The equation of the curve,}\\ y={\displaystyle \int \left(6x^2-12x\right)}dx\\ y=\frac{6x^3}{3}-\frac{12x^2}{2}+c\\ y=2x^3-6x^2+c\\ -14=2{\left(2\right)}^3-6{\left(2\right)}^2+c\text{, at point }P\left(2,-14\right)\\ -14=-8+c\\ c=-6\\ y=2x^3-6x^2-6\end{array}$


(b)
$\begin{array}{l}\frac{dy}{dx}=6x^2-12x\\ \text{At turning points, }\frac{dy}{dx}=0\\ 6x^2-12x=0\\ 6x\left(x-2\right)=0\\ x=0\text{, }x=2\\ \\ x=0,y=2{\left(0\right)}^3-6{\left(0\right)}^2-6=-6\\ x=2,y=2{\left(2\right)}^3-6{\left(2\right)}^2-6=-14\\ \\ \frac{d^{2}y}{d{x}^2}=12x-12\\ \\ \text{When }x=0\text{, }\frac{d^{2}y}{d{x}^2}=12\left(0\right)-12=-12<0\\ \left(0,-6\right)\text{ is a maximum point}\text{.}\\ \\ \text{When }x=2\text{, }\frac{d^{2}y}{d{x}^2}=12\left(2\right)-12=12>0\\ \left(2,-14\right)\text{ is a minimum point}\text{.}\end{array}$