# How do you locate the absolute extrema of the function #f(x)=x^3-12x# on the closed interval [0,4]?

##### 1 Answer

#### Answer:

The absolute extrema of a function,

#### Explanation:

So consider:

The zeros of

So the only critical number in the interval is

The minimum and maximum must occur at one of the values

To finish, evaluate

The minimum is

The maximum is

To understand what we've done it may help to see the graph:

graph{y=(x^3-12x)*sqrt(4-(x-2)^2)/(sqrt(4-(x-2)^2)) [-52.2, 40.28, -23.87, 22.4]}