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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
Answer: The values of c are \(\approx \pm\frac{10}{\sqrt{3}}\) and lie in the interval (-10, 10).

Step by step solution

01

Verify the conditions of the Mean Value Theorem

In this case, the function \(f(x) = x^3\) is a polynomial function, which is continuous and differentiable at all real numbers. Since it is continuous and differentiable on the entire real number line, it is also continuous and differentiable on the closed interval [-10,10].
02

Calculate the average rate of change

The average rate of change (AROC) of a function on an interval [a, b] is the slope of the secant line connecting points a and b, and can be found using the formula: $$\text{AROC}=\frac{f(b)-f(a)}{b-a}$$ Here, a = -10 and b = 10. So, we have: $$\text{AROC}=\frac{f(10)-f(-10)}{10-(-10)}$$ Substitute the function \(f(x) = x^3\) to calculate the AROC: $$\text{AROC}=\frac{(10^3)-((-10)^3)}{10-(-10)}=\frac{2000}{20}=100$$
03

Find the derivative of the function

To find the instantaneous rate of change at any point, we need to calculate the derivative of the given function with respect to x: $$f'(x)=\frac{d}{dx}(x^3)= 3x^2$$
04

Solve for the values of c where the instantaneous rate of change equals the average rate of change

Now, we will set the derivative equal to the AROC and solve for c: $$3x^2= 100$$ To solve for x, we can divide both sides by 3: $$x^2 = \frac{100}{3}$$ Now, take the square root of both sides: $$x = \pm\sqrt{\frac{100}{3}} = \pm\frac{10}{\sqrt{3}}$$ So, the points c where the conclusion of the Mean Value Theorem holds for \(f(x)=x^{3}\) on the interval [-10,10] are \(c \approx \pm\frac{10}{\sqrt{3}}\).

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