Question

How do you find all critical point and determine the min, max and inflection given `f(x)=2x^3-x^2+1`?

Answer 1

`f(x)` has a local maximum for `x=0`, a local minimum for `x=1/3` and an inflection point at `x=1/6`

Explanation:

We can find the critical points by equating the first derivative to zero:

`f(x) = 2x^3-x^2+1`

`f'(x) = 6x^2-2x =2x(3x-1)`

so the critical points are:

`x_1 = 0`
`x_2 = 1/3`

To determine if they are local extrema and to find inflection points we calculate the second derivative:

`f''(x) = 12x-2 =2(6x-1)`

So we have a single inflection point at `x=1/6` where `f(x)` changes from concave down to concave up.

Based on the sign of the second derivative we can also conclude that `x_1` is a local maximum and `x_2` a local minimum.

graph{2x^3-x^2+1 [-0.3227, 0.9273, 0.7125, 1.3375]}