Find the minimum and maximum value of the function y=x3−3x2+6. Find the values of y at which it occurs.
Minima at x=2,minimum value = 6
Given y=x3−3x2+6
Differentiating y w.r.t. 'x', dydx=3x2−6x
Putting dydx=0, we will get the values at which function is maximum or minimum
⇒ 3x2−6x=0⇒x(3x−6)=0
⇒x=0,+2
To distinguish values of x as the point of maximum or minimum, we need 2nd derivative of the function.
∴d2ydx2=6x−6; Now(d2ydx2)x=0=−6<0
∴At x=0⇒It is maximum(d2ydx2)x=2=6(2)−6=6>0∴At x=2⇒It is minimum
Hence x=0 is a point of maximum and x = 2 is a point of minimumSo, maximum value of y = 03−3.0+6=6minimum value of y=(2)3−3(2)2+6=2