#Mark the correct alternative in each of the following
The least value of the function f(x) = x3 – 18x2 + 96x in the interval [0, 9] is
f(x)=x3-18x2+96x
f(x)=3x2-36x+96
f’(x)=0
3x2-36x+96=0
x2-12x+32=0
x2-8x-4x+32=0
x(x-8)-4(x-8)=0
Now we have f(0)=0,f(4)=160,f(9)=135
Hence in the given interval least value is f(0)=0
Option(D)