 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/7/a/i/solution

For

0 \leq x \leq 7

, we have

2 \leq 2 + x \leq 9

. Taking the nonnegative square root gives

G ([0, 7]) = [2, 3] \subseteq [0, 7] .

G^{'} (x) = \frac{1}{2 2 + x},

∣ G^{'} (x) ∣ \leq \frac{1}{2 2} < 1 (0 \leq x \leq 7) .

In particular, the mean value theorem makes

G

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