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Given $f(x) = x + \frac{1}{x}$, we need to find its derivative $f'(x)$. Differentiating with respect to $x$, we get: $$f'(x) = \frac{d}{dx}(x + \frac{1}{x}) = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$$ $$f'(x) = 1 + (-1)x^{-2} = 1 - \frac{1}{x^2}$$
We need to show that $f'(x) > 0$ for $x \ge 1$. $$f'(x) = 1 - \frac{1}{x^2}$$ Since $x \ge 1$, we have $x^2 \ge 1$. Therefore, $\frac{1}{x^2} \le 1$. $$1 - \frac{1}{x^2} \ge 0$$ Thus, $f'(x) \ge 0$ for $x \ge 1$.
Since $f'(x) \ge 0$ for $x \ge 1$, the function $f(x)$ is an increasing function for $x \ge 1$.
Final Answer: f(x) is an increasing function for x ≥ 1
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