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 following function and interval.\n\\( f ( x ) = 4 x ^ { 2 } + 4 x - 3, \\quad - 2, - 1 \\)\nthe value(s) of c that satisfy the equation \\( \\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \\prime } ( c ) \\) is/are \\( \\square \\).\n(type a simplified fraction. 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 following function and interval.\n\\( f ( x ) = 4 x ^ { 2 } + 4 x - 3, \\quad - 2, - 1 \\)\nthe value(s) of c that satisfy the equation \\( \\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \\prime } ( c ) \\) is/are \\( \\square \\).\n(type a simplified fraction. use a comma to separate answers as needed.)

Answer

Explanation:

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

Given (a=-2), (b = - 1), (f(x)=4x^{2}+4x - 3) (f(a)=f(-2)=4\times(-2)^{2}+4\times(-2)-3=16 - 8 - 3=5) (f(b)=f(-1)=4\times(-1)^{2}+4\times(-1)-3=4 - 4 - 3=-3)

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

(\frac{f(-1)-f(-2)}{-1-(-2)}=\frac{-3 - 5}{-1 + 2}=\frac{-8}{1}=-8)

Step3: Find (f^{\prime}(x)) and then solve (f^{\prime}(c))

Differentiate (f(x)=4x^{2}+4x - 3) using the power rule ((x^{n})^\prime=nx^{n - 1}) (f^{\prime}(x)=8x + 4), so (f^{\prime}(c)=8c + 4)

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

Set (8c+4=-8) Subtract (4) from both sides: (8c=-8 - 4=-12) Divide both sides by (8): (c=\frac{-12}{8}=-\frac{3}{2})

Answer:

(-\frac{3}{2})