Question

How do you find all local maximum and minimum points given `y=3x^4-4x^3`?

Answer 1

`x=1` is a local minimum.
`x=0` is an inflection point.

Explanation:

First we find the critical values for `f(x)` equating the first derivative to zero:

`f'(x) =12x^3-12x^2 = 0`

`12x^2(x-1) = 0 => x_1=0, x_2=1` are the critical points.

Now we calculate the second derivative:

`f''(x) = 36x^2-24x = 12x(3x-2)`

So we have `f''(1) = 12 >0` and `x_2=1` is a local minimum.
Instaed we have `f''(0) = 0`, so to determine the nature of `x_1=0` we can look at the sign of the first derivative and we can see that:

`f'(x) = 12x^2(x-1) < 0` around `x=0`, so that `x_1`cannot be a local maximum or minimum and is instead an inflection point.

graph{3x^4-4x^3 [-2.813, 2.812, -1.406, 1.407]}