If f(x) = x + frac{1}{x},x neq 0,then the loc | vedclass

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(B) Given $f(x) = x + \frac{1}{x}$,$x 
eq 0$. Differentiating with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2}$. For local maxima or minima,we s

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$-2$ and $2$

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(B) Given $f(x) = x + \frac{1}{x}$,$x 
eq 0$. Differentiating with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2}$. For local maxima or minima,we set $f'(x) = 0$. $1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1, -1$. Now,we check the sign of $f'(x)$ around these points: For $x  0$. For $-1 The local maximum value is $f(-1) = -1 + \frac{1}{-1} = -2$. For $0  1$,$f'(x) > 0$. Since $f'(x)$ changes from negative to positive at $x = 1$,$x = 1$ is a point of local minima. The local minimum value is $f(1) = 1 + \frac{1}{1} = 2$. Thus,the local maximum value is $-2$ and the local minimum value is $2$.

If $f(x) = x + \frac{1}{x}$,$x \neq 0$,then the local maximum and minimum values of the function $f$ are respectively....

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