Question

How do you find local maximum value of f using the first and second derivative tests: `f(x) = 4x^3 + 3x^2 - 6x + 1`?

Answer 1

`f''(-1)=24*(-1)+6=(-18)` - local maximum
`f''(1/2)=24*(1/2)+6=18` - local minimum

Explanation:

`f'(x)=12*x^2+6*x-6`
`f'(x) =0`
`12x^2+6x-6=0`
`x_1=(-1)`
`x_2=1/2`

there are two values -local maximum `x_1 , x_2`

`f''(x)=24x+6`

`f''(-1)=24*(-1)+6=(-18)` - local maximum, <0
`f''(1/2)=24*(1/2)+6=18` - local minimum, >0