Science:Math Exam Resources/Courses/MATH100/December 2018/Question 07/Solution 1

From UBC Wiki

We are being asked to show that the global maximum of is at most equal to and its global minimum is at least equal to . For this, note that the function is periodic with period and therefore suffices to find the global maximum and the global minimum on the closed interval

First determine the critical points in the interval by computing the derivative and setting it equal to 0:

so implies that . The only solutions of thid equation in the interval are and . It remains to check the value of at the endpoints of the interval and at the two critical points:

Hence, , as desired.