Question

given: y= f(x) = x^3 - 3x +2 determine dy/dx, the second derivative, turning points, min and max, point of inflection

Answer 1

f'(x)=dy/dx=3x^2-3; when 3x^2-3=0 there is a turning-point (the gradient is horizontal). So x^2=1 and x=1 and -1.

Second derivative f"(x)=6x. At x=1 f"(1)>0 (minimum) and at x=-1 f"(-1)<0 (maximum).

When x=1 y=1-3+2=0; when x=-1 y=-1+3+2=4.

The coords for min are (1,0) and max (-1,4).