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Chapter 6: Problem 1

For each of the following functions, verify that the conditions of the Mean Value Theorem are satisfied, and find a value for \(c\) that satisfies the conclusion of the theorem: (a) \(f(x)=x^{3}+2 x, \quad x \in[-2,2]\); (b) \(f(x)=e^{x}, \quad x \in[0,3]\).

Short Answer

Expert verified
For (a) \( c = \pm \frac{2\sqrt{3}}{3} \); for (b) \( c = \ln\left(\frac{e^3 - 1}{3}\right) \).

Step by step solution

01

Verify conditions for Mean Value Theorem in Part (a)

The Mean Value Theorem (MVT) states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). For the function \( f(x) = x^3 + 2x \), it is a polynomial, which means it is continuous and differentiable everywhere, including the interval \([-2, 2]\). Therefore, the conditions of the MVT are satisfied.
02

Apply the Mean Value Theorem to find c for Part (a)

Calculate \( f(-2) \) and \( f(2) \):\[ f(-2) = (-2)^3 + 2(-2) = -8 - 4 = -12 \]\[ f(2) = (2)^3 + 2(2) = 8 + 4 = 12 \]Apply the MVT formula:\[ f'(x) = 3x^2 + 2 \]Set \( f'(c) \) equal to the average rate of change:\[ 3c^2 + 2 = \frac{12 - (-12)}{2 - (-2)} = \frac{24}{4} = 6 \]Solve for \( c \):\[ 3c^2 + 2 = 6 \]\[ 3c^2 = 4 \]\[ c^2 = \frac{4}{3} \]\[ c = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3} \]Since \( c \) must be within \((-2, 2)\), both values \( \frac{2\sqrt{3}}{3} \) and \(-\frac{2\sqrt{3}}{3}\) are valid.
03

Verify conditions for Mean Value Theorem in Part (b)

The function \( f(x) = e^x \) is an exponential function, which is continuous and differentiable on any interval. Specifically, it's continuous on \([0, 3]\) and differentiable on \((0, 3)\), so the MVT conditions are satisfied for this function.
04

Apply the Mean Value Theorem to find c for Part (b)

Calculate \( f(0) \) and \( f(3) \):\[ f(0) = e^0 = 1 \]\[ f(3) = e^3 \]Apply the MVT formula:\[ f'(x) = e^x \]Set \( f'(c) \) equal to the average rate of change:\[ e^c = \frac{e^3 - 1}{3 - 0} = \frac{e^3 - 1}{3} \]Solve for \( c \):Taking natural log both sides:\[ c = \ln\left(\frac{e^3 - 1}{3}\right) \]The value \( c \) should satisfy this expression, and it falls within the interval \((0,3)\).

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Key Concepts

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

Continuity
Continuity is a fundamental concept in calculus and essential for applying the Mean Value Theorem. A function is considered continuous on an interval if there are no breaks, holes, or jumps in its graph within that interval. For a function like a polynomial, such as the one given by \(f(x) = x^3 + 2x\), continuity is inherently satisfied. This is because polynomials are composed of terms with whole number powers of \(x\), which means they can be smoothly traced from left to right without interruption or a point where the function does not exist.
In contrast, discontinuous functions might have gaps or jumps making the application of the Mean Value Theorem impossible. In terms of exponential functions, like \(f(x) = e^x\), they're also known for their smooth and uninterrupted growth or decay, ensuring they are continuous on any interval.
  • Continuity ensures that there are no sudden changes in the value of a function.
  • Both polynomials and exponential functions are naturally continuous over all real numbers.
  • This characteristic is pivotal in meeting the prerequisites for calculus concepts like the Mean Value Theorem.
Differentiability
Differentiability means that a function has a well-defined tangent line at every point within an interval, resulting in a derivative. This property is crucial for the Mean Value Theorem because it requires not just continuity but also that rates of change (slopes of tangents) are calculable.
For the polynomial function \(f(x) = x^3 + 2x\), differentiability is straightforward because the power rule applies, making the derivative easy to find as \(f'(x) = 3x^2 + 2\). Similarly, for the exponential function \(f(x) = e^x\), the derivative is simply \(f'(x) = e^x\), reflecting the self-replicating nature of exponential rates of change.
  • Differentiability confirms that tangent slopes exist at all points in an interval.
  • Having a derivative (like \(f'(x)\)) that is continuous often implies the function itself is differentiable.
  • For both polynomial and exponential functions here, differentiability is naturally satisfied.
Polynomial Functions
Polynomial functions, like \(f(x) = x^3 + 2x\), are expressions involving powers of \(x\). They're among the most straightforward types of functions because they are defined for all real numbers. This universal definition means they're both continuous and differentiable everywhere, making them the perfect candidates for applications of the Mean Value Theorem.
These functions are easy to analyze due to their predictable structure. Simplifying or finding the derivative involves basic power rules, further simplifying calculus operations. In this exercise, the derivative \(f'(x) = 3x^2 + 2\) helps locate the point \(c\) in the interval using the Mean Value Theorem.
  • Polynomial functions are easy to manipulate in calculus, making them educationally beneficial.
  • They inherently satisfy the conditions for continuity and differentiability.
  • The clear formulaic nature facilitates finding derivatives, such as in applying \(f'(c)\) for theorem conclusions.
Exponential Functions
Exponential functions, such as \(f(x) = e^x\), exhibit rapid growth rates and are a cornerstone in calculus. Their constant rate of change, represented by \(e^x\) itself, lends them a unique property where the derivative is the same as the original function. This self-similarity makes exponential functions particular and interesting for both mathematical theory and practical applications.
For the Mean Value Theorem, exponential functions' continuous and differentiable character simplifies the process of verifying theorem conditions and calculating derivatives.
In this exercise, to apply the theorem for \(f(x) = e^x\), we find the derivative as \(f'(x) = e^x\), and then set it equal to the average rate of change over the interval \([0, 3]\), easily calculating the point \(c\) using logarithmic transformations, further showcasing their analytical tractability.
  • Exponential functions grow or decay at rates proportional to their current value.
  • They are always both continuous and differentiable, making calculus applications straightforward.
  • The Mean Value Theorem application is simplified due to easy derivative calculations, maintaining function shape.

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Most popular questions from this chapter

Prove that the following functions \(f\) are not differentiable at the given point \(c\) : (a) \(f(x)=|x|^{\frac{1}{2}}, x \in \mathbb{R}, c=0\); (b) \(f(x)=[x], x \in \mathbb{R}, c=1\). Here \([x]\) denotes the integer Looking back at Example 2 , it looks as though chords joining the origin to points \((h, f(h))\) have slopes that tend to a limit 1 if \(h \rightarrow 0^{+}\), whereas they have slopes that tend to a limit \(-1\) if \(h \rightarrow 0^{-}\). This suggests the concept of one-sided derivatives that will be useful in our work later on. Definitions Let \(f\) be defined on an interval \(I\), and \(c \in I .\) Then the left derivative of \(f\) at \(c\) is $$ f_{L}^{\prime}(c)=\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c} \text { All these definitions are } $$ analogous to definitions for continuity that you met in provided that this limit exists. In this case, we say that \(f\) is differentiable on the left at \(c\). Similarly, the right derivative of \(f\) at \(c\) is $$ f_{R}^{\prime}(c)=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \text { or } f_{R}^{\prime}(c)=\lim _{h \rightarrow 0^{+}} Q(h), $$ provided that this limit exists. In this case, we say that \(f\) is differentiable on the right at \(c\). A function \(f\) whose domain contains an interval \(I\) is differentiable on \(I\) if it is differentiable at each interior point of \(I\), differentiable on the right at the left end-point of \(I\) (if this belongs to \(I\) ) and differentiable on the left at the right end-point of \(I\) (if this belongs to \(I\) ). | {:[," Theorem 1 "],[,c" as an interior point is differentiable at "c" if and only if "f" is both differenti- "],[," able on the left at "c" and differentiable on the right at "c" AND "],[,f_(L)^(')(c),=f_(R)^(')(c).]:} | We omit a proof of this straight-forward result. The common value is simply f^(')(c). | | :--- | :--- | $$ \begin{aligned} &=\frac{\left\\{(-1+h)+(-1+h)^{2}\right\\}-\left\\{(-1)+(-1)^{2}\right\\}}{h} \\\ &=\frac{-h+h^{2}}{h} \\ &=-1+h \rightarrow-1 \quad \text { as } h \rightarrow 0^{+} \end{aligned} $$ It follows that \(f\) is differentiable on the right at \(-1\), and \(f_{R}^{\prime}(-1)=-1\) At 0, the function is defined on either side of 0 , but by a different formula; we therefore examine each side separately. At 0, for \(0

For the function \(f(x)=x^{3}-3 x^{2}+1, x \in \mathbb{R}\), determine those points \(c\) where \(f^{\prime}(c)=0\). Using the Second Derivative Test, determine whether these correspond to local maxima, local minima or neither.

Prove the following inequalities: (a) \(x \geq \sin x\), for \(x \in\left[0, \frac{1}{2} \pi\right]\) (b) \(\frac{2}{3} x+\frac{1}{3} \geq x^{2}\), for \(x \in[0,1]\).

Verify that the conditions of Rolle's Theorem are satisfied by the function $$ f(x)=x^{4}-4 x^{3}+3 x^{2}+2, \quad x \in[1,3] $$ and determine a value of \(c\) in \((1,3)\) for which \(f^{\prime}(c)=0\).

Find the derivative of each of the following functions: (a) \(f(x)=\tan x, \quad x \in \mathbb{R}-\left\\{\pm \frac{1}{2} \pi, \pm \frac{3}{2} \pi, \pm \frac{5}{2} \pi, \ldots\right\\}\) (b) \(f(x)=\operatorname{cosec} x, \quad x \in \mathbb{R}-\\{0, \pm \pi, \pm 2 \pi, \ldots\\}\) (c) \(f(x)=\sec x, \quad x \in \mathbb{R}-\left\\{\pm \frac{1}{2} \pi, \pm \frac{3}{2} \pi, \pm \frac{5}{2} \pi, \ldots\right\\}\) (d) \(f(x)=\cot x, \quad x \in \mathbb{R}-\\{0, \pm \pi, \pm 2 \pi, \ldots\\} .\)
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