 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/5/a/ii/solution

The extreme value theorem guarantees extrema on the closed interval. Compare its endpoints and the stationary points

0, \pm 10/3

f (\pm 3) = 330, f (0) = 6, f (\pm 10/3) = - \frac{338}{27} .

Thus

[- 3, 3] arg max f = {- 3, 3}, [- 3, 3] arg min f = {- \frac{10}{3}, \frac{10}{3}} .

The two-point answer in each case also follows from the even function symmetry.

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