Question

How do you find the critical points for `f(x) = x^3 - 15x^2 + 4` and the local max and min?

Answer 1

The critical points are: `x_1=0` and `x_2=10`.
`x_1` is a maximum, while `x_2` is a minimum.

Explanation:

To identify the critical points, we have to solve the equation:

`f'(x)=0`

that is:

`3x^2-30x=0`

`3x(x-10) =0`

so that the critical points for `f(x)` are `x_1=0` and `x_2=10`.
We can now calculate the second derivative:

`f''(x) = 6x-30`

and note that around `x_1` `f''(x)` is negative, therefore `x_1` is a maximum, while around `x_2` it is positive, therefore `x_2` is a minimum.

graph{x^3-15x^2+4 [-80, 80, -640, 640]}