Questions: The geometric mean of two positive numbers (a) and (b) is the number (sqrtab). Show that the value of (c) in the conclusion the Mean Value Theorem for (f(x)=frac1x) on an interval of positive numbers ([a, b]) is (c=sqrtab). Apply the Mean Value Theorem to the given function. [ fracf(b)-f(a)b-a = fracfrac1b-frac1ab-a = -frac1ab ] Simplify. By definition, this expression for (fracf(b)-f(a)b-a) is equal to (f'(c)). Using (f(x)=frac1x), find another expression for (f'(c)). [ f(x)=frac1x ] [ f'(c)=-frac1c^2 quad textEvaluate the derivative of f(x) text at x=c ] Use the two expressions found previously for (f'(c)) to write an equation to solve for (c). (Type an equation.)

Solution Steps

To solve this problem, we need to apply the Mean Value Theorem (MVT) to the function \( f(x) = \frac{1}{x} \) over the interval \([a, b]\). The MVT states that there exists some \( c \) in the interval \((a, b)\) such that the derivative at \( c \), \( f'(c) \), is equal to the average rate of change of the function over \([a, b]\). We will first calculate the average rate of change, then find the derivative of \( f(x) \), and finally set these two expressions equal to solve for \( c \).

The average rate of change of the function \( f(x) = \frac{1}{x} \) over the interval \([a, b]\) is given by:

\[ \frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a} = -\frac{1}{a b} \]

The derivative of the function \( f(x) = \frac{1}{x} \) is:

\[ f'(x) = -\frac{1}{x^2} \]

Evaluating this at \( x = c \) gives:

\[ f'(c) = -\frac{1}{c^2} \]

According to the Mean Value Theorem, we set the average rate of change equal to the derivative at \( c \):

\[ -\frac{1}{c^2} = -\frac{1}{a b} \]

Rearranging the equation gives:

\[ \frac{1}{c^2} = \frac{1}{a b} \]

Taking the reciprocal of both sides leads to:

\[ c^2 = a b \]

Taking the square root of both sides results in:

\[ c = \sqrt{a b} \]

Final Answer