For what values of x is the function f(x) = x | vedclass

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(B) Given the function $f(x) = x + \frac{1}{x}$.To find where the function is strictly increasing,we calculate the derivative $f'(x)$ and set $f'(x) >

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$|x| > 1$

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(B) Given the function $f(x) = x + \frac{1}{x}$.To find where the function is strictly increasing,we calculate the derivative $f'(x)$ and set $f'(x) > 0$.$f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.For strictly increasing,$f'(x) > 0$,so $1 - \frac{1}{x^2} > 0$.$\frac{x^2 - 1}{x^2} > 0$.Since $x^2 > 0$ for all $x 
eq 0$,the inequality holds if $x^2 - 1 > 0$.$x^2 > 1$,which implies $|x| > 1$.

For what values of $x$ is the function $f(x) = x + \frac{1}{x}, x \neq 0$ strictly increasing?

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