consider the function ( f(x)=2x^{3}-12x^{2}-72x + 10 ) on the interval ( -6,7 ). find the average or mean slope of the function on this interval. by the mean value theorem, we know there exists a ( c ) in the open interval ( (-6,7) ) such that ( f(c) ) is equal to this mean slope. for this problem, there are two values of ( c ) that work. the smaller one is and the larger one is question help: video message instructor

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Accepted Answer

# Explanation:
## Step1: Calculate the average slope
The formula for the average slope (mean value) of a function $y = f(x)$ on the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$.
Here, $a=-6$, $b = 7$, $f(x)=2x^{3}-12x^{2}-72x + 10$.
First, find $f(-6)$:
$$
\begin{align*}
f(-6)&=2\times(-6)^{3}-12\times(-6)^{2}-72\times(-6)+10\
&=2\times(-216)-12\times36 + 432+10\
&=-432-432 + 432+10\
&=-422
\end{align*}
$$
Next, find $f(7)$:
$$
\begin{align*}
f(7)&=2\times7^{3}-12\times7^{2}-72\times7+10\
&=2\times343-12\times49-504 + 10\
&=686-588-504 + 10\
&=-406
\end{align*}
$$
Then, the average slope $\frac{f(7)-f(-6)}{7-(-6)}=\frac{-406-(-422)}{13}=\frac{16}{13}\approx1.23$. But wait, let's do it more accurately:
$$
\begin{align*}
\frac{f(7)-f(-6)}{7 + 6}&=\frac{(2\times7^{3}-12\times7^{2}-72\times7 + 10)-(2\times(-6)^{3}-12\times(-6)^{2}-72\times(-6)+10)}{13}\
&=\frac{2(7^{3}+ 6^{3})-12(7^{2}+6^{2})-72(7 - 6)}{13}\
&=\frac{2(343 + 216)-12(49 + 36)-72}{13}\
&=\frac{2\times559-12\times85-72}{13}\
&=\frac{1118-1020 - 72}{13}\
&=\frac{26}{13}=2
\end{align*}
$$

## Step2: Find the derivative of $f(x)$
Differentiate $f(x)=2x^{3}-12x^{2}-72x + 10$ using the power rule $(x^{n})^\prime=nx^{n - 1}$.
$f^\prime(x)=6x^{2}-24x-72$

## Step3: Set $f^\prime(c)$ equal to the average slope
Since $f^\prime(c)=2$ (the average slope we found), we have the equation $6c^{2}-24c-72 = 2$.
Rearrange it to $6c^{2}-24c-74 = 0$, or $3c^{2}-12c-37=0$.
Using the quadratic formula $c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$ (here $a = 3$, $b=-12$, $c=-37$):
$$
\begin{align*}
c&=\frac{12\pm\sqrt{(-12)^{2}-4\times3\times(-37)}}{2\times3}\
&=\frac{12\pm\sqrt{144+444}}{6}\
&=\frac{12\pm\sqrt{588}}{6}\
&=\frac{12\pm14\sqrt{3}}{6}\
& = 2\pm\frac{7\sqrt{3}}{3}\approx2\pm4.04
\end{align*}
$$
The two solutions are $c=2-\frac{7\sqrt{3}}{3}\approx - 2.04$ and $c=2+\frac{7\sqrt{3}}{3}\approx6.04$

# Answer:
The average slope is $2$. The smaller value of $c$ is $2-\frac{7\sqrt{3}}{3}\approx - 2.04$ and the larger value of $c$ is $2+\frac{7\sqrt{3}}{3}\approx6.04$

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Question

Explanation:

Step1: Calculate the average slope

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

Step3: Set \(f^\prime(c)\) equal to the average slope

Answer: