Question

find the value or values of c that satisfy the equation \\( \frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c ) \\) in the conclusion of the mean value theorem for the given function and interval.
\\( f ( x ) = \sqrt { x - 3 }, 3,9 \\)
\\( c = \square \\)
(simplify your answer. use a comma to separate answers as needed.)

Explanation:

Step1: Calculate \( f(a) \) and \( f(b) \)

Given \( a = 3 \), \( b = 9 \), \( f(x)=\sqrt{x - 3}\).
\( f(a)=f(3)=\sqrt{3 - 3}=0 \)
\( f(b)=f(9)=\sqrt{9 - 3}=\sqrt{6} \)

Step2: Calculate \(\frac{f(b)-f(a)}{b - a}\)

\(\frac{f(9)-f(3)}{9 - 3}=\frac{\sqrt{6}-0}{6}=\frac{\sqrt{6}}{6}\)

Step3: Find \( f^{\prime}(x) \)

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \( f(x)=(x - 3)^{\frac{1}{2}}\), \( f^{\prime}(x)=\frac{1}{2}(x - 3)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x - 3}}\)

Step4: Set \( f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\) and solve for \( c \)

\(\frac{1}{2\sqrt{c - 3}}=\frac{\sqrt{6}}{6}\)
Cross - multiply: \(6 = 2\sqrt{6}\sqrt{c - 3}\)
Divide both sides by \(2\sqrt{6}\): \(\sqrt{c - 3}=\frac{6}{2\sqrt{6}}=\frac{\sqrt{6}}{2}\)
Square both sides: \(c-3=\frac{6}{4}=\frac{3}{2}\)
Add 3 to both sides: \(c=\frac{3}{2}+3=\frac{3 + 6}{2}=\frac{9}{2}\)

Answer:

\(\frac{9}{2}\)