Find the coordinates of the turning points of the curve y = 4/3 x^3 + 3x^2-4x+1
First differentiating by the rule that xn differentiates to nxn-1 we have that dy/dx = 4x2+6x-4.
At the turning points of a curve the differential is equal to 0 so we set 0=dy/dx = 4x2+6x-4, by factorising we can see that 0= 2(2x2+3x-2) = 2(2x-1)(x+2), so our turning points are at x=1/2 and x=-2 as this is when dy/dx =0.
to find the coordinates for these points we plug the x values into the original equation y=4/3 x3+3x2-4x+1 and we find that the turning points are (1/2, -1/12) and (-2, 31/3).
TE
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