 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/6/b/i/solution

The derivative factors as

g^{'} (x) = 4 x^{3} - 24 x^{2} + 36 x = 4 x (x - 3)^{2} .

The polynomial is differentiable everywhere, so its critical points occur precisely where this derivative vanishes. Hence

x = 0 and x = 3.

Their values are

g (0) = 5

and

g (3) = 32

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