If f(x) = x + frac{1}{x} for x 0,then its g | vedclass

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(D) Given the function $f(x) = x + \frac{1}{x}$ for $x > 0$.First,we find the derivative of the function with respect to $x$:$f'(x) = \frac{d}{dx}(x +

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None of these

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(D) Given the function $f(x) = x + \frac{1}{x}$ for $x > 0$.First,we find the derivative of the function with respect to $x$:$f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.To find the critical points,we set $f'(x) = 0$:$1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = \pm 1$.Since the domain is $x > 0$,we only consider $x = 1$.Now,we check the second derivative to determine the nature of the critical point:$f''(x) = \frac{d}{dx}(1 - x^{-2}) = 2x^{-3} = \frac{2}{x^3}$.At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.Since $f''(1) > 0$,the function has a local minimum at $x = 1$,and the minimum value is $f(1) = 1 + \frac{1}{1} = 2$.As $x 	o 0^+$,$f(x) 	o \infty$,and as $x 	o \infty$,$f(x) 	o \infty$.Therefore,the function does not have a greatest value (maximum) on the interval $(0, \infty)$.Thus,the correct option is $(d)$.

If $f(x) = x + \frac{1}{x}$ for $x > 0$,then its greatest value is:

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