Question

How do you find the local extrema for `f(x) = 2x^3 - x^2 - 4x +3`?

Answer 1

`f(x)` has a local maximum for `x=-2/3` and a local minimum for `x=1`

Explanation:

You find the critical values for `f(x)` by identifying where the first derivative equals zero:

`f'(x) = 6x^2-2x-4 = 0`

`x= frac(1+-sqrt(1+24)) 6= frac (1+-5) 6`,

so: `x_1=-2/3` and `x_2=1`

Now, as `f(x)` is a second degree polynomial with leading positive coefficient, we know that:

`f'(x) > 0` for `x in (-oo,x_1)` and `x in (x_2,+oo)`
`f'(x)<0` for `x in (x_1,x_2)`

so `x_1` is a maximum and `x_2` is a minimum.

graph{2x^3-x^2-4x+3 [-10, 10, -5, 5]}