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Chapter 4: Problem 3

3\. Prove: If \(f(x)\) is defined for \(a

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Question: Prove that if a function \(f(x)\) has a continuous derivative over the interval \((a,b)\) and the magnitude of its derivative is bounded by some constant \(K\), i.e., \(\left|f^{\prime}(x)\right| \leq K\), then \(f(x)\) is uniformly continuous over the interval \((a,b)\). Answer: By using the mean value theorem, we can relate the derivative of \(f(x)\) to the absolute difference of \(f(x)\) and \(f(y)\). Then, utilizing the boundedness property of the derivative, we can show that for all \(x\) and \(y\) in \((a,b)\), \(|f(x) - f(y)| \leq K|y - x|\). Finally, we can choose \(\delta = \frac{\epsilon}{K}\) to satisfy the definition of uniform continuity, proving that \(f(x)\) is uniformly continuous over the interval \((a,b)\).

Step by step solution

01

Recall the definition of uniformly continuous

A function \(f(x)\) is uniformly continuous over an interval if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x\) and \(y\) in the interval with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \epsilon\). We need to show that the given conditions on the derivative of \(f(x)\) imply this definition of uniformly continuous.
02

Apply the Mean Value Theorem

The mean value theorem states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a point \(c \in (a, b)\) such that the average rate of change of the function (\(f^{\prime}(c)\)) is equal to the average change of the function itself, i.e., \(\frac{f(b) - f(a)}{b - a} = f^{\prime}(c)\). Our goal is to use the mean value theorem to relate the derivative of \(f(x)\) to the absolute difference of \(f(x)\) and \(f(y)\).
03

Relate derivative to the difference of function values

Let \(x,y \in (a,b)\) such that \(|x-y|<\delta\). We define \(x\) and \(y\) such that \(y>x\). We can apply the mean value theorem on the interval \([x,y]\) since \(f(x)\) is defined on the interval \((a,b)\) and \(f^\prime(x)\) is continuous on the same interval. According to the mean value theorem, there exists a \(c \in (x, y)\), such that: \(\frac{f(y) - f(x)}{y - x} = f^{\prime}(c)\) Multiplying both sides by \(|y-x|\): \(|f(y) - f(x)|=|f^{\prime}(c)||y - x|\)
04

Utilize the boundedness property of the derivative

We are given that the derivative of the function \(f\) is bounded such that \(\left|f^{\prime}(x)\right| \leq K\). Using this inequality and since \(f^{\prime}(c)\) is a derivative of \(f(x)\), we have: \(\left|f^{\prime}(c)\right|\leq K\) Then, from the result in Step 3: \(|f(y) - f(x)|\leq K|y - x|\)
05

Show that \(f(x)\) is uniformly continuous

We want to show that \(f(x)\) is uniformly continuous over \((a,b)\). According to the definition of uniformly continuous, for any \(\epsilon>0\), we need to find a \(\delta>0\) such that for all \(x\) and \(y\) in \((a,b)\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \epsilon\). As we know that the function's derivative is bounded, we can choose our \(\delta=\frac{\epsilon}{K}\) (since \(K>0\)). Then, we have: \(|f(y) - f(x)|\leq K|y - x| < K\delta = K\frac{\epsilon}{K}=\epsilon\) Thus, for all \(x\) and \(y\) in \((a,b)\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \epsilon\). Therefore, \(f(x)\) is uniformly continuous over the interval \((a,b)\).

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Key Concepts

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

Mean Value Theorem
When exploring the behavior of functions, the Mean Value Theorem (MVT) is a vital tool. It provides a solid link between the average rate of change over an interval and the instantaneous rate of change at a particular point in that interval.
Given a function \(f(x)\), if it is continuous over a closed interval \([a, b]\) and differentiable over the open interval \((a, b)\), then MVT assures us that there is at least one point \(c\) in \((a, b)\) where the derivative \(f'(c)\) is exactly equal to the 鈥渁verage鈥� derivative across the whole interval:
\[\frac{f(b) - f(a)}{b - a} = f'(c)\]
This equation means the slope of the tangent at \(c\) mirrors the slope of the line connecting \(f(a)\) and \(f(b)\). In practical terms, if you consider any smooth slope鈥攁 hill, for example鈥攖he MVT guarantees a spot where the slope matches the overall climb, illustrating the precise 'nature' of average rates in calculus.
  • This theorem helps prove important properties in calculus such as uniform continuity and integral evaluation.
  • It relies on the function being both continuous and differentiable, reinforcing the importance of these concepts in analysis.
Bounded derivative
A function is said to have a bounded derivative if the values of its derivative do not exceed a certain fixed bound throughout its domain. In mathematical terms, if you have a function \(f(x)\) where \(f'(x)\) is bounded, it means there exists a constant \(K\) such that for every value of \(x\) in the interval, \(|f'(x)| \leq K\).
The notion of bounded derivatives is crucial because it provides a control over how steep or flat the function can get, thus ensuring that changes in the function's values are not too drastic.
  • This condition is pivotal in establishing uniform continuity of functions, since it implies a consistent rate of change.
  • Bounded derivatives lead to predictions about the smoothness and behavior of a function.

Consider driving at a uniform speed limit; if a road鈥檚 slope change rate was bounded, you could anticipate a smooth drive without abrupt steepness.
The exercise's solution uses this concept to show that the bounded nature of the derivative allows for selecting a suitable \(\delta\) against any \(\epsilon\) in the uniform continuity proof.
Continuous function
Continuous functions are those that allow you to draw their graph without lifting your pencil. In more formal terms, a function \(f(x)\) is continuous at a point \(c\) if the limit of \(f(x)\) as \(x\) approaches \(c\) is equal to \(f(c)\). For a function to be continuous over an interval, this property must hold at every point throughout the interval.
Continuity is a foundational concept in calculus, and it ensures that there are no breaks, jumps, or holes in the graph of a function through the interval considered.
  • It facilitates the application of the Mean Value Theorem, as the graph must be smooth enough to compute slopes between points accurately.

  • Continuous functions on closed intervals exhibit the Intermediate Value Property, which can be especially useful in solving equations.

Understanding continuity is indispensable in analyzing functions鈥� behavior, forming the base for deeper calculus concepts such as differentiability and integrability.
In the exercise, the fact that \(f(x)\) is continuous ensures that the conditions of the Mean Value Theorem are met, allowing further investigation into the uniform continuity of the function.

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Most popular questions from this chapter

Transform to cylindrical coordinates but do not evaluate: a) \(\iiint_{R_{x z}} x^{2} y d x d y d z\), where \(R_{x y z}\) is the region \(x^{2}+y^{2} \leq 1,0 \leq z \leq 1\); b) \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{1+x+y}\left(x^{2}-y^{2}\right) d z d y d x\).

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Let an ellipse be given by parametric equations: \(x=a \cos \phi, y=b \sin \phi, b>a>0\). Show that the length of arc from \(\phi=0\) to \(\phi=\alpha\) is given by $$ s=b \int_{0}^{\alpha} \sqrt{1-k^{2} \sin ^{2} \phi} d \phi, \quad k^{2}=\frac{b^{2}-a^{2}}{b^{2}} . $$ 5\. Show that the function \(F(x)\) defined by (4.24) has the following properties: a) \(F(x)\) is defined and continuous for all \(x\); b) as \(x\) increases, \(F(x)\) increases; c) \(F(x+\pi)-F(x)=2 K\), where \(K\) is a constant; d) \(\lim _{x \rightarrow \infty} F(x)=\infty, \lim _{x \rightarrow-\infty} F(x)=-\infty\).
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