Question

consider the function (f(x) = 2x^3 - 6x^2 - 48x + 4) on the interval (-4, 10). 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 ((-4, 10)) 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

Explanation:

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<reasoning>

Calculate the average slope on the interval

\[

$$\begin{aligned} a &= -4,\quad b = 10 \\ f(x) &= 2x^3 - 6x^2 - 48x + 4 \\ f(-4) &= 2(-4)^3 - 6(-4)^2 - 48(-4) + 4 \\ &= 2(-64) - 6(16) + 192 + 4 \\ &= -128 - 96 + 192 + 4 = -28 \\ f(10) &= 2(10)^3 - 6(10)^2 - 48(10) + 4 \\ &= 2(1000) - 6(100) - 480 + 4 \\ &= 2000 - 600 - 480 + 4 = 924 \\ \text{Average Slope} &= \frac{f(10) - f(-4)}{10 - (-4)} \\ &= \frac{924 - (-28)}{14} \\ &= \frac{952}{14} = 68 \end{aligned}$$

\]

Find the derivative of the function

\[

$$\begin{aligned} f'(x) &= \frac{d}{dx}(2x^3 - 6x^2 - 48x + 4) \\ &= 6x^2 - 12x - 48 \end{aligned}$$

\]

Solve for c where the derivative equals the average slope

\[

$$\begin{aligned} f'(c) &= 68 \\ 6c^2 - 12c - 48 &= 68 \\ 6c^2 - 12c - 116 &= 0 \\ 3c^2 - 6c - 58 &= 0 \\ c &= \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-58)}}{2(3)} \\ &= \frac{6 \pm \sqrt{36 + 696}}{6} \\ &= \frac{6 \pm \sqrt{732}}{6} \\ &= 1 \pm \frac{\sqrt{732}}{6} \\ &= 1 \pm \frac{2\sqrt{183}}{6} \\ &= 1 \pm \frac{\sqrt{183}}{3} \\ c_1 &= 1 - \frac{\sqrt{183}}{3} \approx -3.509 \\ c_2 &= 1 + \frac{\sqrt{183}}{3} \approx 5.509 \end{aligned}$$

\]
</reasoning>

<answer>
Consider the function \(f(x) = 2x^3 - 6x^2 - 48x + 4\) on the interval \([-4, 10]\). Find the average or mean slope of the function on this interval.
<blank>68</blank>

By the Mean Value Theorem, we know there exists a \(c\) in the open interval \((-4, 10)\) 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 <blank>\(1 - \frac{\sqrt{183}}{3}\)</blank>

and the larger one is <blank>\(1 + \frac{\sqrt{183}}{3}\)</blank>
</answer>

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Answer:

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<reasoning>

Calculate the average slope on the interval

\[

$$\begin{aligned} a &= -4,\quad b = 10 \\ f(x) &= 2x^3 - 6x^2 - 48x + 4 \\ f(-4) &= 2(-4)^3 - 6(-4)^2 - 48(-4) + 4 \\ &= 2(-64) - 6(16) + 192 + 4 \\ &= -128 - 96 + 192 + 4 = -28 \\ f(10) &= 2(10)^3 - 6(10)^2 - 48(10) + 4 \\ &= 2(1000) - 6(100) - 480 + 4 \\ &= 2000 - 600 - 480 + 4 = 924 \\ \text{Average Slope} &= \frac{f(10) - f(-4)}{10 - (-4)} \\ &= \frac{924 - (-28)}{14} \\ &= \frac{952}{14} = 68 \end{aligned}$$

\]

Find the derivative of the function

\[

$$\begin{aligned} f'(x) &= \frac{d}{dx}(2x^3 - 6x^2 - 48x + 4) \\ &= 6x^2 - 12x - 48 \end{aligned}$$

\]

Solve for c where the derivative equals the average slope

\[

$$\begin{aligned} f'(c) &= 68 \\ 6c^2 - 12c - 48 &= 68 \\ 6c^2 - 12c - 116 &= 0 \\ 3c^2 - 6c - 58 &= 0 \\ c &= \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-58)}}{2(3)} \\ &= \frac{6 \pm \sqrt{36 + 696}}{6} \\ &= \frac{6 \pm \sqrt{732}}{6} \\ &= 1 \pm \frac{\sqrt{732}}{6} \\ &= 1 \pm \frac{2\sqrt{183}}{6} \\ &= 1 \pm \frac{\sqrt{183}}{3} \\ c_1 &= 1 - \frac{\sqrt{183}}{3} \approx -3.509 \\ c_2 &= 1 + \frac{\sqrt{183}}{3} \approx 5.509 \end{aligned}$$

\]
</reasoning>

<answer>
Consider the function \(f(x) = 2x^3 - 6x^2 - 48x + 4\) on the interval \([-4, 10]\). Find the average or mean slope of the function on this interval.
<blank>68</blank>

By the Mean Value Theorem, we know there exists a \(c\) in the open interval \((-4, 10)\) 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 <blank>\(1 - \frac{\sqrt{183}}{3}\)</blank>

and the larger one is <blank>\(1 + \frac{\sqrt{183}}{3}\)</blank>
</answer>

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