 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/6/c/solution

f (1) = 1/2

f^{'} (1) = - 1/2

and

f^{''} (1) = 1/2

. Therefore

f_{2} (x) = \frac{1}{2} - \frac{1}{2} (x - 1) + \frac{1}{4} (x - 1)^{2} = \frac{x ^{2} - 4 x + 5}{4} .

The quadratic coefficient is

f^{''} (1) /2!

, not

f^{''} (1)

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