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

1 Answer
Jul 10, 2015

Answer:

#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#