Let #p(x)=x^3-9x^2+10# Find the absolute maximum and minimum value of #p# on the interval #[-2,8]#?

1 Answer
Mar 11, 2017

Absolute maximum: #(0, 10)#
Absolute minimum: #(6, -98)#

Explanation:

We start by differentiating.

#p'(x) = 3x^2 - 18x#

We now find the critical numbers, which occur when the derivative is #0# or is undefined. Since polynomials are continuous, we will only have critical numbers when #dy/dx = 0#.

#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: #x = -1#

#p(-1) = 3(-1)^2 - 18(-1) = 3 + 18 = 21#

This means that #p(x)# is increasing on #(-oo, 0)#. Before confirming that #x= 0# is an absolute maximum though we must check the y-values of the function at the end points.

#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, #(0, 10)# is an absolute maximum on #[-2, 8]# (always specify, because on #(-oo, oo)#, this function has no absolute maximum).

Finally, we must ensure that #x = 6# is an absolute minimum.

Test Point #x = 5#

#p(5) = 5^3 - 9(5)^2 + 10#

#p(5) = -90#

We can now see that #p(x)# is decreasing on #(0, 6)#. This will mean that we have a minimum of some sort at #x = 6#. We know for sure that it will be either an absolute or local minimum because this critical point has a derivative equal to #0#, so the function changes vertical direction, or the derivative goes from decreasing to increasing.

#p(6)= 6^3 - 9(6)^2 + 10#

#p(6) = -98#

Since this is smaller than both #p(-2)# and #p(8)#, we can see that this is an absolute minimum.

Hopefully this helps!