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 - 7 }, 7,9 \\) \\( c = \\) (simplify your answer. use a comma to separate answers as needed.)

Explanation:

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

Given \( f(x)=\sqrt{x - 7} \), \( a = 7 \), \( b = 9 \).
\( f(a)=f(7)=\sqrt{7 - 7}=0 \)
\( f(b)=f(9)=\sqrt{9 - 7}=\sqrt{2} \)

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

\(\frac{f(b)-f(a)}{b - a}=\frac{\sqrt{2}-0}{9 - 7}=\frac{\sqrt{2}}{2}\)

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

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), for \( f(x)=(x - 7)^{\frac{1}{2}} \), \( f^{\prime}(x)=\frac{1}{2}(x - 7)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x - 7}} \)

Step4: Solve \( f^{\prime}(c)=\frac{f(b)-f(a)}{b - a} \)

Set \( \frac{1}{2\sqrt{c - 7}}=\frac{\sqrt{2}}{2} \)
Cross - multiply: \( 2 = 2\sqrt{2}\sqrt{c - 7} \)
Divide both sides by \( 2 \): \( 1=\sqrt{2}\sqrt{c - 7} \)
Square both sides: \( 1 = 2(c - 7) \)
Expand: \( 1=2c-14 \)
Add \( 14 \) to both sides: \( 15 = 2c \)
Solve for \( c \): \( c=\frac{15}{2}=7.5 \)

Answer:

\(7.5\)