**Option : (D)**

f(x) = x^{3 }- 18x^{2 }+ 96x

f(x) = 3x^{2 }- 36x + 96

f’(x) = 0

3x^{2} - 36x + 96 = 0

x^{2 }- 12x + 32 = 0

x^{2} - 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**