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

Dec 21, 2016

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:

$f ' \left(x\right) = 0$

that is:

$3 {x}^{2} - 30 x = 0$

$3 x \left(x - 10\right) = 0$

so that the critical points for $f \left(x\right)$ are ${x}_{1} = 0$ and ${x}_{2} = 10$.
We can now calculate the second derivative:

$f ' ' \left(x\right) = 6 x - 30$

and note that around ${x}_{1}$ $f ' ' \left(x\right)$ 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]}