Let #p(x)=x^3-9x^2+10# Find the absolute maximum and minimum value of #p# on the interval #[-2,8]#?
1 Answer
Absolute maximum:
Absolute minimum:
Explanation:
We start by differentiating.
#p'(x) = 3x^2 - 18x#
We now find the critical numbers, which occur when the derivative is
#0 = 3x(x - 6)#
#x= 0 and 6#
We now test around the points to check whether the derivative is increasing or decreasing.
Test Point:
#p(-1) = 3(-1)^2 - 18(-1) = 3 + 18 = 21#
This means that
#p(-2) = (-2)^3 - 9(-2)^2 + 10#
#p(-2) = -8 -9(4) +10#
#p(-2) = -34#
This works!
#p(8) = 8^3 - 9(8)^2 + 10#
#p(8) = 512 - 576 + 10#
#p(8) = -54#
This works!
Therefore,
Finally, we must ensure that
Test Point
#p(5) = 5^3 - 9(5)^2 + 10#
#p(5) = -90#
We can now see that
#p(6)= 6^3 - 9(6)^2 + 10#
#p(6) = -98#
Since this is smaller than both
Hopefully this helps!