Question

1-the curve y=x^2+A÷x has the gradient of 8 when x =5.calculate the value of A.
2-find the maximum point and the minimum points of :y=2x^3-3x^2-12x+1.

Answer 1

1)

$\frac{dy}{dx}=2x-\frac{A}{x^2}.\\ \frac{dy}{dx}|_{x=5}=8\to 10-\frac{A}{25}=8\to A=50.$

2)

$\frac{dy}{dx}=6x^2-6x-12.\\ \frac{dy}{dx}=0\to 6x^2-6x-12=0\to x=-1,\;\;x=2.\\ \frac{d^2y}{dx^2}=12x-6.\\ \frac{d^2y}{dx^2}|_{x=-1}=-18<0.\\ \frac{d^2y}{dx^2}|_{x=2}=18>0.$

So, the maximum point is $(-1, y(-1))=(-1, 8).$

the minimum point is $(2, y(2))=(2,-19).$