How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3-6x^2+15#?

Question

Luke Alvarez · Accepted Answer

Relative maximum: #(0,15)#
Relative minimum: #(4,-17)# Relative maximums and minimums occur whenever the derivative equals 0. Using the power rule, we find the derivative is: #f'(x)=3x^2-12x# Setting it equal to #0# yields: #0=3x^2-12x# #0=x^2-4x# #0=x(x-4)# #x=0# and #x=4# To find the exact minimum and maximum, we evaluate these two #x#-values (called critical points): #f(0)=(0)^3-6(0)^2+15=15-># maximum #f(4)=(4)^3-6(4)^2+15=64-96+15=-17-># minimum Looking at the graph of the function, we can see that there is indeed a local max at #(0,15)# and a local min at #(4,-17)#. graph{x^3-6x^2+15 [-42.95, 49.5, -19.6, 26.67]}

Luke Alvarez · Answer

undefined To find the exact relative maximum and minimum of the polynomial function f(x) = x^3 - 6x^2 + 15:

1. Take the derivative of the function f(x) to find critical points.
   f'(x) = 3x^2 - 12x

2. Set the derivative equal to zero and solve for x to find critical points.
   3x^2 - 12x = 0
   x(3x - 12) = 0
   x = 0 or x = 4

3. Evaluate the function at the critical points and endpoints to find the y-values.
   f(0) = 15
   f(4) = 15

4. Determine the nature of the critical points using the second derivative test.
   f''(x) = 6x - 12

For x = 0:
   f''(0) = -12
   Since f''(0) < 0, it indicates a relative maximum at x = 0.

For x = 4:
   f''(4) = 12
   Since f''(4) > 0, it indicates a relative minimum at x = 4.

5. Therefore, the exact relative maximum is (0, 15) and the exact relative minimum is (4, 15).