Question

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

Explanation:

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

Given \(a=-3\), \(b = 0\), and \(f(x)=5x^{2}-2x - 3\).
For \(x=a=-3\):
\(f(-3)=5\times(-3)^{2}-2\times(-3)-3=5\times9 + 6-3=45+6 - 3=48\).
For \(x = b=0\):
\(f(0)=5\times0^{2}-2\times0-3=-3\).

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

\(\frac{f(0)-f(-3)}{0-(-3)}=\frac{-3 - 48}{3}=\frac{-51}{3}=-17\).

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

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

Step4: Solve \(f^{\prime}(c)=-17\)

Set \(10c-2=-17\).
Add \(2\) to both sides: \(10c=-17 + 2=-15\).
Divide both sides by \(10\): \(c=\frac{-15}{10}=-\frac{3}{2}\).

Answer:

\(-\frac{3}{2}\)