find the value or values of c that satisfy the equation \\( \\frac { f ( b ) - f ( a ) } { b - a } = f ^ {…

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.\n\\( f ( x ) = \\sqrt { x - 3 }, 3,9 \\)\n\\( c = \\square \\)\n(simplify your answer. use a comma to separate answers as needed.)

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.\n\\( f ( x ) = \\sqrt { x - 3 }, 3,9 \\)\n\\( c = \\square \\)\n(simplify your answer. use a comma to separate answers as needed.)

Answer

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})