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Chapter 4: Problem 6

At what points \(c\) does the conclusion of the Mean Value Theorem hold for \(f(x)=x^{3}\) on the interval [-10,10]\(?\)

Short Answer

Expert verified
The points c at which the Mean Value Theorem holds for the given cubic function are \(c_1 = -\frac{10\sqrt{3}}{3}\) and \(c_2 = \frac{10\sqrt{3}}{3}\).

Step by step solution

01

Determine if MVT conditions are met

A cubic function is both continuous and differentiable on its entire domain, so \(f(x) = x^3\) meets the conditions for the Mean Value Theorem to be applied on the interval [-10, 10].
02

Find the derivative of the function

To find the derivative of \(f(x) = x^3\), we will use the power rule: \(f'(x) = \frac{d}{dx} (x^3) = 3x^2\)
03

Compute the average rate of change

Now, we'll compute the average rate of change of the function over the interval [-10, 10]: \(\frac{f(b) - f(a)}{b - a} = \frac{f(10) - f(-10)}{10 - (-10)} = \frac{(10^3) - (-10)^3}{20} = \frac{1000 - (-1000)}{20} = \frac{2000}{20} = 100\)
04

Set the derivative equal to the average rate of change and solve for c

Next, we'll set the derivative equal to the average rate of change: \(3x^2 = 100\) Now, we'll solve for x: \(x^2 = \frac{100}{3}\) \(x = \pm\sqrt{\frac{100}{3}} = \pm\frac{10}{\sqrt{3}} = \pm\frac{10\sqrt{3}}{3}\) Thus, the Mean Value Theorem holds at the two points: \(c_1 = -\frac{10\sqrt{3}}{3}\) \(c_2 = \frac{10\sqrt{3}}{3}\)

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