Mathematics MCQ - Calculus
f(x)=2x3-15x2+36x+10
f '(x)=6x2-30x+36 and f ''(x)=12x-30
For maxima and minima,
f '(x)=0
6x2-30x+36 = 0 or
x2- 5x + 6 = 0 or
X = 3,2
Putting these values of x in equation (ii), we get
f '(3) = 36 - 30 = + 6 positive
Hence minimum value = 3
f ' (2) = 24 - 30 = - 6 negative, hence maximum value = 2
