 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/5/solution

For

f (x) = x e^{- 2 x^{2}}

f^{'} (x) = e^{- 2 x^{2}} (1 - 4 x^{2}) .

The positive exponential factor makes

f

increasing for

0 < x < 1/2

and decreasing for

1/2 < x < 1

. Its only interior stationary point is therefore a maximum at

x = 1/2

. Checking it together with both endpoints,

f (0) = 0, f (1/2) = \frac{1}{2} e^{- 1/2}, f (1) = e^{- 2} .

Consequently

[0, 1] min f = 0 at x = 0, [0, 1] max f = \frac{1}{2 e} at x = \frac{1}{2} .

Checking the endpoints is necessary for extrema on a closed interval.

 New to topics? Read the docs here!