Find the turning points of the curve y=2x^3 - 3x^2 - 14.
First differentiate the equation: dy/dx = 6x^2 - 6xSet this equal to 0 as at turning points the change in gradient is 0: 0 = 6x^2 - 6x6x(x-1)=06x=0 therefore x=0(x-1)=0 therefore x=1x=1,0Now substitute back into equation 1 to find y values and hence co ordinatesx=1 y=-15x=0 y=-14
ER
Related Maths A Level answers
All answers ▸When finding the turning points of a curve, how can I tell if it is a maximum, minimum or a point of inflection?
Find the equation of the tangent for x = 2cos (2y +pi)
A curve has parametric equations x= 2sin(t) , y= cos(2t) + 2sin(t) for -1/2 π≤t≤ 1/2π , show that dy/dx = - 2sin(t)+ 1
Solve to find sin x , 4cos^2 + 7sin x -7 =0