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The function f(x)=2x^3-15 x^2+36 x+4 is ...

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

A

`x=2`

B

`x=4`

C

`x=0`

D

`x=3`

Text Solution

Verified by Experts

The correct Answer is:
A
Given, `f(x)=2x^(3)-15x^(2)+36x+4`
`impliesf'(x)=6x^(2)-30x+36`
Put f'(x)=0, for maximum or minima.
`therefore(6x^(2)-5x+6)=0`
`implies(x-3)(x-2)=0`
`impliesx=3,2`
Now, `f"(x)=12x-30`
`f"(3)=36-30=6`
and `f"(2)=24-30=-6lt0`
`therefore(f)x` is maximum at x=2.
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