Answer to Question #99386 in Calculus for Nhlaphu Mulaudzi

Question #99386
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.
1
Expert's answer
2019-11-26T11:23:44-0500

1)

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


2)

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


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