 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/4/b/solution

f^{(4)} (x) = e^{x} - e x, f^{(5)} (x) = e^{x} - e .

For

0 < x < 1

, the fifth derivative is negative, so the fourth derivative is strictly decreasing. Since

f^{(4)} (1) = 0

, strict decrease towards that endpoint implies

f^{(4)} (x) > 0

for

x < 1

. On the closed interval, $0 \leq f^{(4)} (x) \leq 1$ , with the maximum at

0

and minimum at

1

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