The largest value of `2x^3-3x^2-12 x+5` for `-2lt=xlt=4` occurs at x equals

Options

A. `-4`
B. `0`
C. `1`
D. `4`

Answer

D

Explanation

Let `f(x)=2x^(3)-3x^(2)-12x+5`
`impliesf"(x)=6x^(2)-6x-12`
For maxima or minima put `f'(x)=0`
`implies6x^(2)-6x-12=0`
`impliesx=-1,2`
Now `f''(x)=12x-6`
`''(-1)=-12-6=-18lt0` maxima
`:.f(x)` is maxima at `x=-1`
At `x=-1, f(X)=-2-3+12+5=12`
At `x=-2, f(x)=-16-12+24+5=1`
At `x=4,f(x)=128-48-48+5=37`
Hence larges value of `f(x)` in the given interval is at the point `x=4`.