The function `f(x)=2x^3-15 x^2+36 x+4` is maximum at `x=` (a) 3 (b) 0 (c) 4 (d) 2

Options

A. `x=2`
B. `x=4`
C. `x=0`
D. `x=3`

Answer

A

Explanation

Given `f(x)=2x^(3)-15x^(2)+36 x+4`
`impliesf'(x)=6x^(2)-30x+36`………….i
For maxima or minima put `f'(x)=0`
`implies6x^(2)-30x+36=0`
`impliesx^(2)-5x+6=0`
`implies(x-2)(x-3)=0impliesx=2,3`
Now `f''(X)=12x-30`
At `x=2, f''(2)=24-30=-6lt0`
Therefore `f(x)` is maximum at `x=2`.