The least value of the sum of any positive re | vedclass

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(B) Let the positive real number be $x$,where $x > 0$.We want to find the minimum value of the function $f(x) = x + \frac{1}{x}$.To find the critical

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$2$

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(B) Let the positive real number be $x$,where $x > 0$.We want to find the minimum value of the function $f(x) = x + \frac{1}{x}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(x) = 0$ gives $1 - \frac{1}{x^2} = 0$,which implies $x^2 = 1$.Since $x$ is a positive real number,we take $x = 1$.Now,we check the second derivative to confirm the minimum:$f''(x) = \frac{2}{x^3}$.At $x = 1$,$f''(1) = 2 > 0$,which confirms that $f(x)$ has a local minimum at $x = 1$.The minimum value is $f(1) = 1 + \frac{1}{1} = 2$.Alternatively,by the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for $x > 0$,$\frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}} = 1$,so $x + \frac{1}{x} \ge 2$.

The least value of the sum of any positive real number and its reciprocal is

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