Question

How do you find the global extreme values for `y=x^2-6x-1` on [-2,2]?

Answer 1

`f_min (2)=-9`

`f_max (-2)=15`

Explanation:

The possible extreme points of such a function are:
1) vertex of the parabola `V(p,q)`
2) end points of the interval (in this case `a=-2, b=2`)

First we calculate the vertex:

`p=-b/(2a)=6/2=3`

3 does not belong to the interval mentioned in the task, so we don't have to calculate `q`, we know, that the extreme values are at the ends of the interval.

So we calculate:

`f(-2)=4+12-1=15`

`f(2)=4-12-1=-9`

Now we can write the answer.

`f_min (2)=-9`

`f_max (-2)=15`