Find the coordinates of the point on the curve y=x^3-8x at which the gradient of the curve is 4?

1 Answer
Jim G.
Nov 28, 2017

#(-2,8)" and "(2,-8)#

Explanation:

#"using "color(blue)"differential calculus"#

#•color(white)(x)m_(color(red)"tangent")=dy/dx#

#y=x^3-8x#

#rArrdy/dx=3x^2-8#

#"since gradient of tangent is given as "m=4#

#rArr3x^2-8=4#

#rArr3x^2-12=0larrcolor(blue)"quadratic equation"#

#rArr3(x^2-4)=0#

#rArr3(x-2)(x+2)=0rArrx=+-2#

#"substitute these values into equation for y-coordinates"#

#x=-2toy=(-2)^3-8(-2)=-8+16=8#

#x=2toy=2^3-16=-8#

#rArr(-2,8)" and "(2,-8)" are the points on the curve"#

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