The function f(x) = x + displaystyle1/x has

The correct answer is Option (1) → local maxima at \(x = -1\)

The critical points are where \(f'(x) = 0\):

    \( \displaystyle f'(x) = 1 - \frac{1}{x^2} \)

    \(1 - \frac{1}{x^2} = 0\)
    \(x^2 = 1,\quad x = \pm 1\)

By the second derivative test:

    \( \displaystyle f''(x) = \frac{2}{x^3} \)

    \(f''(-1) = -2 \lt 0\): local maximum at \(x = -1\)
    \(f''(1) = 2 \gt 0\): local minimum at \(x = 1\)

So f has a local maximum at \(x = -1\) (value \(f(-1) = -2\)) and a local minimum at \(x = 1\) (value \(f(1) = 2\)).

Tip: Here the local maximum value (−2) is smaller than the local minimum value (2). A local maximum is only the largest value near that point, so do not reject a maximum just because its value is smaller.