 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/7/solution

f^{'} (x) = exp (\frac{x}{1 + x ^{2}}) \frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}} .

The exponential function and denominator are positive, so the stationary points occur exactly at

x = \pm 1

. Their values are

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

The derivative is negative for

∣ x ∣ > 1

and positive for

∣ x ∣ < 1

, so

x = - 1

x = 1

a local maximum. They are also global extrema, since

- 1/2 \leq x / (1 + x^{2}) \leq 1/2

 New to topics? Read the docs here!