question 2
5 pts
find the number c satisfying the conclusion of the mean value theorem for the function $f(x)=\frac{1}{3}x^{2}-x^{2}-3x + 5$, where $xin-3,0$.
$c=-1$
$c = 1-sqrt{7}$
$c=1+sqrt{7}$
$c = 3$

Question

Accepted Answer

# Explanation:
## Step1: Recall Mean - Value Theorem formula
The Mean - Value Theorem states that if $y = f(x)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then $f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}$, where $a=-3$, $b = 0$ and $f(x)=\frac{1}{3}x^{3}-x^{2}-3x + 5$. First, find $f(a)$ and $f(b)$.
$$
\begin{align*}
f(-3)&=\frac{1}{3}(-3)^{3}-(-3)^{2}-3(-3)+5\
&=\frac{1}{3}\times(-27)-9 + 9+5\
&=-9 - 9+9 + 5\
&=-4
\end{align*}
$$
$$
\begin{align*}
f(0)&=\frac{1}{3}(0)^{3}-(0)^{2}-3(0)+5\
&=5
\end{align*}
$$
Then $\frac{f(0)-f(-3)}{0-(-3)}=\frac{5-(-4)}{3}=\frac{9}{3}=3$.

## Step2: Find the derivative of $f(x)$
Differentiate $f(x)=\frac{1}{3}x^{3}-x^{2}-3x + 5$ using the power - rule $(x^{n})^\prime=nx^{n - 1}$.
$$
\begin{align*}
f^{\prime}(x)&=\frac{d}{dx}(\frac{1}{3}x^{3}-x^{2}-3x + 5)\
&=\frac{1}{3}\times3x^{2}-2x-3\
&=x^{2}-2x - 3
\end{align*}
$$

## Step3: Set $f^{\prime}(c)$ equal to $\frac{f(0)-f(-3)}{0 - (-3)}$ and solve for $c$
Set $x^{2}-2x - 3=3$, which can be rewritten as $x^{2}-2x-6 = 0$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-2$, $c=-6$).
$$
\begin{align*}
c&=\frac{2\pm\sqrt{(-2)^{2}-4\times1\times(-6)}}{2\times1}\
&=\frac{2\pm\sqrt{4 + 24}}{2}\
&=\frac{2\pm\sqrt{28}}{2}\
&=\frac{2\pm2\sqrt{7}}{2}\
&=1\pm\sqrt{7}
\end{align*}
$$
Since $c\in[-3,0]$, we take $c = 1-\sqrt{7}$.

# Answer:
$c = 1-\sqrt{7}$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Mean - Value Theorem formula

Step2: Find the derivative of \(f(x)\)

Step3: Set \(f^{\prime}(c)\) equal to \(\frac{f(0)-f(-3)}{0 - (-3)}\) and solve for \(c\)

Answer: