 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/1/solution

Use completing the square after taking out the leading coefficient:

3 (x^{2} - 4 x) + 11 = 3 [(x - 2)^{2} - 4] + 11 = 3 (x - 2)^{2} - 1.

Thus

a = 3

b = - 2

and

c = - 1

. Since a real square is nonnegative, the global minimum is $- 1$ , attained at $x = 2$ .

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