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Given the function $f(x) = x + \frac{1}{x}$. To find the critical points, we differentiate with respect to $x$: $$f'(x) = 1 - \frac{1}{x^2}$$
Set $f'(x) = 0$: $$1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1, -1$$
Find the second derivative $f''(x)$: $$f''(x) = \frac{2}{x^3}$$ For $x = 1$: $f''(1) = 2 > 0$, so $x = 1$ is a point of local minima. The value is $f(1) = 1 + 1 = 2$. For $x = -1$: $f''(-1) = -2 < 0$, so $x = -1$ is a point of local maxima. The value is $f(-1) = -1 - 1 = -2$.
Local minimum value is $2$. Local maximum value is $-2$. Since $2 > -2$, the statement "local minimum value < local maximum value" is false. However, checking the options provided, we see that (C) states the local maximum value is $-2$, which is correct.
Final Answer: C