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 following function and interval. \\( f ( x ) = 5 x ^ { 2 } - 2 x - 3 \\), \\( - 1,0 \\) the value(s) of c that satisfy the equation \\( \frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c ) \\) is/are \\( \square \\). (type a simplified fraction. use a comma to separate answers as needed.)

Explanation:

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

Given \( a=-1\), \(b = 0\), and \(f(x)=5x^{2}-2x - 3\).
For \(x=a=-1\):
\(f(-1)=5\times(-1)^{2}-2\times(-1)-3=5 + 2-3=4\).
For \(x = b=0\):
\(f(0)=5\times0^{2}-2\times0-3=-3\).

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

\(\frac{f(0)-f(-1)}{0-(-1)}=\frac{-3 - 4}{1}=-7\).

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

Differentiate \(f(x)=5x^{2}-2x - 3\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(f^{\prime}(x)=10x-2\), so \(f^{\prime}(c)=10c-2\).

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

Set \(10c-2=-7\).
Add \(2\) to both sides: \(10c=-7 + 2=-5\).
Divide both sides by \(10\): \(c=-\frac{5}{10}=-\frac{1}{2}\).

Answer:

\(-\frac{1}{2}\)