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(C) Let the function be $f(x) = x + \frac{1}{x}$.To find the critical points,we find the derivative $f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(x) = 0$,we

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(C) Let the function be $f(x) = x + \frac{1}{x}$.To find the critical points,we find the derivative $f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(x) = 0$,we get $1 = \frac{1}{x^2}$,which implies $x^2 = 1$,so $x = \pm 1$.Now,we find the second derivative $f''(x) = \frac{2}{x^3}$.For $x = 1$,$f''(1) = 2 > 0$,so the function has a local minimum at $x = 1$.The minimum value is $f(1) = 1 + \frac{1}{1} = 2$.For $x = -1$,$f''(-1) = -2 < 0$,so the function has a local maximum at $x = -1$,where $f(-1) = -2$.

What is the minimum value of the sum of a real number $x$ and its reciprocal?

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