 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/3/solution

f^{'} (x) = - 2 x exp (2 - x^{2}) .

Since

x^{2} \geq 0

, the exponent is at most

2

, with equality precisely at

x = 0

. The exponential function is increasing, so

x \in R max f (x) = e^{2}, attained at x = 0.

The sign of the derivative, positive for

x < 0

and negative for

x > 0

, gives the same conclusion.

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