The local maximum value of f(x) = x + frac{1} | vedclass

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(A) To find the local maximum value of $f(x) = x + \frac{1}{x}$,we first find the derivative $f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(x) = 0$,we get $1

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-$2$

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(A) To find the local maximum value of $f(x) = x + \frac{1}{x}$,we first find the derivative $f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(x) = 0$,we get $1 = \frac{1}{x^2}$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$.We use the second derivative test: $f''(x) = \frac{2}{x^3}$.For $x = 1$,$f''(1) = 2 > 0$,so $x = 1$ is a local minimum.For $x = -1$,$f''(-1) = \frac{2}{(-1)^3} = -2 The local maximum value is $f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$.

The local maximum value of $f(x) = x + \frac{1}{x}$ for $x < 0$ is equal to:

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