the function ( f(x)=x^{3}+14 x^{2}+56 x + 74 ) is graphed below. plot a line segment connecting the points on ( f ) where ( x=-9 ) and ( x=-4 ). afterwards, determine all values of ( c ) which satisfy the conclusion of the mean value theorem for ( f ) on the closed interval ( -9 leq x leq -4 ). plot a line by clicking in two locations. click the line to delete it.

Question

Accepted Answer

# Explanation:
## Step1: Calculate $ f(-9) $ and $ f(-4) $
$$
\begin{align*}
f(-9)&=(-9)^{3}+14(-9)^{2}+56(-9)+74\
&=-729 + 14\times81-504 + 74\
&=-729+1134-504 + 74\
&=-25
\end{align*}
$$
$$
\begin{align*}
f(-4)&=(-4)^{3}+14(-4)^{2}+56(-4)+74\
&=-64+14\times16-224 + 74\
&=-64 + 224-224+74\
&=10
\end{align*}
$$
## Step2: Find the slope of the secant line
The slope $ m=\frac{f(-4)-f(-9)}{-4-(-9)}=\frac{10 - (-25)}{-4 + 9}=\frac{35}{5}=7 $
## Step3: Find the derivative of $ f(x) $
$ f^{\prime}(x)=3x^{2}+28x + 56 $
## Step4: Set $ f^{\prime}(c)=7 $
$$
\begin{align*}
3c^{2}+28c + 56&=7\
3c^{2}+28c+49&=0
\end{align*}
$$
Using the quadratic formula $ c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} $ with $ a = 3 $, $ b=28 $, $ c = 49 $
$$
\begin{align*}
c&=\frac{-28\pm\sqrt{28^{2}-4\times3\times49}}{2\times3}\
&=\frac{-28\pm\sqrt{784 - 588}}{6}\
&=\frac{-28\pm\sqrt{196}}{6}\
&=\frac{-28\pm14}{6}
\end{align*}
$$
$$
c_{1}=\frac{-28 + 14}{6}=\frac{-14}{6}=-\frac{7}{3}, \quad c_{2}=\frac{-28-14}{6}=\frac{-42}{6}=-7
$$
Since $ -9\leq c\leq -4 $, we check the values. $ c=-7 $ is in the interval $ [-9,-4] $ and $ c =-\frac{7}{3}\approx - 2.33$ is not in the interval $ [-9,-4] $

# Answer:
$ c=-7 $

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate \( f(-9) \) and \( f(-4) \)

Step2: Find the slope of the secant line

Step3: Find the derivative of \( f(x) \)

Step4: Set \( f^{\prime}(c)=7 \)

Answer: