From the Mean Value Theorem,f(b) - f(a) = (b | vedclass

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(A) Given $f(x) = \frac{1}{x}$.Then $f'(x) = -\frac{1}{x^2}$.According to the Mean Value Theorem,$f'(x_1) = \frac{f(b) - f(a)}{b - a}$.Substituting th

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$\sqrt{ab}$

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(A) Given $f(x) = \frac{1}{x}$.Then $f'(x) = -\frac{1}{x^2}$.According to the Mean Value Theorem,$f'(x_1) = \frac{f(b) - f(a)}{b - a}$.Substituting the values,we get $-\frac{1}{x_1^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a}$.Simplifying the right side: $\frac{\frac{a - b}{ab}}{b - a} = \frac{-(b - a)}{ab(b - a)} = -\frac{1}{ab}$.Thus,$-\frac{1}{x_1^2} = -\frac{1}{ab}$,which implies $x_1^2 = ab$.Since $a < x_1 < b$,we have $x_1 = \sqrt{ab}$.

From the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(x_1)$ where $a < x_1 < b$. If $f(x) = \frac{1}{x}$,then $x_1 = $

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