Let's say that the function is y=f(x)=2x3−21x2+36x−20∴f′(x)=dxdf(x)=dxd(2x3−21x2+36x−20)=6x2−42x+36 And, f′(x)=dx2d2f(x)=dxd[dxdf(x)]=dxd(6x2−42x+36)=12x−42 For maxima\/minima points, f′(x)=0. ⇒6x2−42x+36=0⇒x2−7x+6=0⇒x2−6x−x+6=0⇒x(x−6)−(x−6)=0⇒(x−6)(x−1)=0⇒x−6=0 or x−1=0⇒x=6 OR x=1 Now, let's check these points for maxima\/minima by inspecting the values of f′′(x) at these points. f′(6)=12(6)−42=72−42=30.f′(1)=12(1)−42=12−42=−30. Since, f′(6)=30>0, it is the point of minimum value. And the minimum value is f(6) : =2(6)3−21(6)2+36(6)−20=432−756+216−20=−128