 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/6/solution

y^{'} = e^{- x} (3 x^{2} - x^{3}) = x^{2} e^{- x} (3 - x) .

For

x > 0

, the derivative is positive before

3

and negative after

3

. Also

y (0) = 0

and

x^{3} e^{- x} \to 0

x \to \infty

. Thus the global maximum on the requested range is

x = 3, y_{m a x} = 27 e^{- 3} .

 New to topics? Read the docs here!