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Chapter 3: Problem 37

Determine whether the Mean Value Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If the Mean Value Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). If the Mean Value Theorem cannot be applied, explain why not. \(f(x)=x^{2}, \quad[-2,1]\)

Short Answer

Expert verified
Yes, the Mean Value Theorem can be applied to the function \(f(x) = x^2\) on the interval \([-2,1]\). The value of \(c\) that satisfies the Mean Value Theorem is \(c = -0.5\).

Step by step solution

01

Examining the conditions for MVT

We have the function \(f(x) = x^2\). It's a polynomial function, which means it's continuous and differentiable for all real numbers. The function is continuous and differentiable both on the closed interval \([-2, 1]\) and open interval \((-2, 1)\). Therefore, MVT can be applied.
02

Applying Mean Value Theorem

Next, apply the MVT by setting up the equation \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). Here, the function \(f(x)=x^2\), so its derivative \(f^{\prime}(x)\) is \(2x\). Hence the equation becomes, \(2c = \frac{f(1) - f(-2)}{1 - (-2)}\).
03

Solving for c

Substitute \(f(1)\) and \(f(-2)\) in the equation. Thus, \(2c = \frac{1^2 - (-2)^2}{1 - (-2)}\). Calculate the right side, giving \(2c = -1\). Solve for \(c\), thus \(c = -0.5\).
04

Verifying the solution

Since the value of \(c\) we found is within the open interval \((-2, 1)\), the solution satisfies the conditions of the Mean Value Theorem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is an area of mathematics that deals with rates of change and accumulations of quantities. It's split into two main branches: differential calculus, which focuses on the derivative of functions, and integral calculus, which looks at finding the area under curves. Understanding calculus is crucial for solving a range of problems in not just mathematics, but also other sciences and engineering. In the context of the Mean Value Theorem (MVT), calculus comes into play as it provides a way to relate the average rate of change of a function over an interval to the instantaneous rate of change at a particular point.
Derivative
The derivative is a fundamental concept in differential calculus. It represents the rate at which a quantity changes. To put it simply, the derivative of a function at a point is the slope of the tangent line to the function's graph at that point. It's typically denoted as \(f'(x)\) or \(\frac{d}{dx}f(x)\). Derivatives can help predict how a function will behave and change, which is invaluable in fields like physics and economics. In the Mean Value Theorem exercise given, finding the derivative of the function \(f(x) = x^2\) is essential in solving for the value of \(c\) that satisfies the theorem's conditions.
Continuity
Continuity in calculus is about whether a function's graph has any breaks, holes, or jumps. If you can draw the graph of the function without lifting your pen, it's considered continuous. More formally, a function is continuous at a point if the function's value at that point is the limit of the function as you approach the point from both the left and the right. For the Mean Value Theorem to apply, the function in question must be continuous on the closed interval from \(a\) to \(b\). In the given exercise, the function \(f(x) = x^2\) is continuous everywhere, which is particularly important for applying the MVT.
Differentiability
Differentiability refers to whether a function has a derivative at every point in an interval. When a function is differentiable, it's smooth and has no sharp corners or cusps. A differentiable function is also continuous, but the reverse isn't always true. For the Mean Value Theorem to work, the function should not only be continuous on a closed interval \[a, b\] but also differentiable on the open interval \(a, b\). The function \(f(x) = x^2\) in the exercise is differentiable as it has a derivative, specifically \(f'(x) = 2x\), which exists for all real \(x\). This meets one of the critical conditions to apply the MVT and find the values of \(c\) within the interval.

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Most popular questions from this chapter

Volume and Surface Area The measurement of the edge of a cube is found to be 15 inches, with a possible error of 0.03 inch. (a) Use differentials to approximate the possible propagated (a) Use differentials the volume of the cube. (b) Use differentials to approximate the possible propagated error in computing the surface area of the cube. (c) Approximate the percent errors in parts (a) and (b).

Find the minimum value of $$\frac{(x+1 / x)^{6}-\left(x^{6}+1 / x^{6}\right)-2}{(x+1 / x)^{3}+\left(x^{3}+1 / x^{3}\right)}\( for \)x>0$$

Numerical, Graphical, and Analytic Analysis An exercise room consists of a rectangle with a semicircle on each end. A 200 -meter running track runs around the outside of the room. (a) Draw a figure to represent the problem. Let \(x\) and \(y\) represent the length and width of the rectangle. (b) Analytically complete six rows of a table such as the one below. (The first two rows are shown.) Use the table to guess the maximum area of the rectangular region. $$ \begin{array}{|c|c|c|}\hline \text { Length, } x & {\text { Width, } y} & {\text { Area, } x y} \\ \hline 10 & {\frac{2}{\pi}(100-10)} & {(10) \frac{2}{\pi}(100-10) \approx 573} \\ \hline 20 & {\frac{2}{\pi}(100-20)} & {(20) \frac{2}{\pi}(100-20) \approx 1019} \\ \hline\end{array} $$ (c) Write the area \(A\) as a function of \(x\) . (d) Use calculus to find the critical number of the function in part (c) and find the maximum value. (e) Use a graphing utility to graph the function in part (c) and verify the maximum area from the graph.

True or False? In Exercises \(47-50\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(y=f(x), f\) is increasing and differentiable, and \(\Delta x>0\) , then \(\Delta y \geq d y\)

Comparing Functions In Exercises 83 and \(84,\) (a) use a graphing utility to graph \(f\) and \(g\) in the same viewing window, (b) verify algebraically that \(f\) and \(g\) represent the same function, and (c) zoom out sufficiently far so that the graph appears as a line. What equation does this line appear to have? (Note that the points at which the function is not continuous are not readily seen when you zoom out.) $$ \begin{array}{l}{f(x)=-\frac{x^{3}-2 x^{2}+2}{2 x^{2}}} \\ {g(x)=-\frac{1}{2} x+1-\frac{1}{x^{2}}}\end{array} $$
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