The function f(x) = x + frac{1}{x}, (x neq 0) | vedclass

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(A) Let $f(x) = x + \frac{1}{x}$.Differentiating with respect to $x$,we get:$f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$.For the function to be n

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$[-1, 1]$

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(A) Let $f(x) = x + \frac{1}{x}$.Differentiating with respect to $x$,we get:$f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$.For the function to be non-increasing,we must have $f'(x) \le 0$.$\frac{x^2 - 1}{x^2} \le 0$.Since $x^2 > 0$ for all $x 
eq 0$,the condition simplifies to $x^2 - 1 \le 0$.$x^2 \le 1$,which implies $x \in [-1, 1]$.However,the function is undefined at $x = 0$. Thus,the function is non-increasing on the intervals $[-1, 0)$ and $(0, 1]$.Given the options,the interval $[-1, 1]$ (excluding $0$) is the correct domain where the derivative is non-positive.

The function $f(x) = x + \frac{1}{x}, (x \neq 0)$ is a non-increasing function in the interval

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