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

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

The first three derivatives factor as

f^{'} (x) = 2 x^{3} (3 x^{2} - 10), f^{''} (x) = 30 x^{2} (x^{2} - 2), f^{'''} (x) = 120 x (x^{2} - 1) .

Let

a = 10/3

. Sign analysis of the factors gives

f^{'} (x) > 0 f^{''} (x) > 0 f^{'''} (x) > 0 ⟺ x \in (- a, 0) \cup (a, \infty), ⟺ x \in (- \infty, - 2) \cup (2, \infty), ⟺ x \in (- 1, 0) \cup (1, \infty) .

All displayed endpoints are excluded because a factor vanishes there; the extra zero of

f^{''}

at

0

does not create a positive interval.

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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