The real number x when added to its inverse g | vedclass

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(B) Let the function be $f(x) = x + \frac{1}{x}$ for $x > 0$.To find the minimum value,we find the derivative $f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(

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(B) Let the function be $f(x) = x + \frac{1}{x}$ for $x > 0$.To find the minimum value,we find the derivative $f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(x) = 0$,we get $1 - \frac{1}{x^2} = 0$,which implies $x^2 = 1$. Since $x > 0$,we have $x = 1$.Now,we check the second derivative $f''(x) = \frac{2}{x^3}$.At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.Since the second derivative is positive,the function $f(x)$ has a local minimum at $x = 1$.

The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to

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