/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Let \(f\) be a continuous functi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(f\) be a continuous function whose domain includes the interval \([a, b]\) and let \(x_{0} \in(a, b)\) have \(f\left(x_{0}\right)>0\). Show that \(f\) is positive throughout some interval surrounding \(x_{0}\).

Short Answer

Expert verified
Function \( f \) is positive in the interval \((x_0 - \delta, x_0 + \delta)\).

Step by step solution

01

Understanding the Problem

We need to show that if a continuous function \( f \) has a positive value at some point \( x_0 \) within an interval \((a, b)\), then \( f \) must be positive in some smaller interval around \( x_0 \). Essentially, we are looking to apply the properties of continuous functions.
02

Apply the Definition of Continuity

Since \( f \) is continuous at \( x_0 \), for any \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( |x - x_0| < \delta \), then \( |f(x) - f(x_0)| < \epsilon \). In particular, choose \( \epsilon = f(x_0) \), which is positive.
03

Choose the Interval

Because \( |f(x) - f(x_0)| < f(x_0) \), it implies \( -f(x_0) < f(x) - f(x_0) < f(x_0) \). By rearranging, we get \( 0 < f(x) < 2f(x_0) \). Thus, \( f(x) \) is positive in the interval \((x_0 - \delta, x_0 + \delta)\).
04

Conclusion

Therefore, there exists an interval \((x_0 - \delta, x_0 + \delta)\) such that \( f(x) > 0 \) for all \( x \) within this interval. This shows \( f \) is positive in some neighborhood around \( x_0 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Intermediate Value Theorem
The Intermediate Value Theorem (IVT) is a fundamental principle in calculus that applies to continuous functions. It states that if a function \( f \) is continuous on a closed interval \([a, b]\), and \( f(a) \) and \( f(b) \) are of opposite signs, there must be at least one point \( c \) in the interval \( (a, b) \) where \( f(c) = 0 \). This theorem is incredibly useful because it assures us that a continuous function will take on every value between \( f(a) \) and \( f(b) \) at some point within the interval.

Applying this to our exercise, we don't have the scenario of encountering a zero directly, but it supports the idea that continuous functions don't "jump" and must transition smoothly, which suggests steady behavior, like the positivity observed around \( x_0 \).

If \( f(x_0) > 0 \), IVT reassures us that \( f \) won't suddenly drop below zero in its neighborhood unless there's another point balancing it out to zero—a situation which the continuity prevents in a sufficiently small interval around \( x_0 \).
Epsilon-Delta Definition
The epsilon-delta definition is the formal definition of a limit, crucial for understanding continuity. A function \( f(x) \) is said to be continuous at a point \( x_0 \) if, for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( |x - x_0| < \delta \), then \( |f(x) - f(x_0)| < \epsilon \).

In practical terms, for any tiny deviation \( \epsilon \) we allow in the function's output, there is a corresponding narrow band \( \delta \) near \( x_0 \) where the function's outputs deviate by less than \( \epsilon \).
  • This concept reinforces the smoothness of continuous functions.
  • It helps us isolate a segment, or interval, around \( x_0 \) where behavior like maintaining positivity is guaranteed.
In our exercise, selecting \( \epsilon = f(x_0) \) ensures that \( f \) stays positive in the interval since its deviations are limited both above and below the reference value \( f(x_0) \).
Interval
An interval is a set of numbers lying between two endpoints. In mathematics, we often distinguish intervals by being open or closed.

Closed intervals \([a, b]\) include their endpoints while open intervals \((a, b)\) do not. For our purpose, the concepts of intervals are crucial because analyzing the behavior of functions often involves sections of their domain.

In the exercise, we look at \((x_0 - \delta, x_0 + \delta)\), which constitutes a small open interval surrounding \( x_0 \).
  • Choosing an interval is key to localize the behavior of \( f \).
  • This localization helps confirm that within this smaller scope, \( f \) remains positive.
Understanding how these segments come together to prove properties of \( f \) helps form a complete picture of its behavior.
Neighborhood
In calculus, a neighborhood around a point is a set of points surrounding it within an interval. If we say "neighborhood of \( x_0 \)," it implies there is some interval \((x_0 - \delta, x_0 + \delta)\) that covers the points near \( x_0 \).

In solving problems related to continuous functions, neighborhoods allow us to analyze behavior not just at a single point, but in a small region around it.

In the context of our exercise, using neighborhoods helps establish where exactly the function \( f \) keeps its positivity near \( x_0 \).
  • These neighborhoods, resultant from the epsilon-delta conditions, ensure a zone of consistent function behavior.
  • They are particularly helpful in visualizing and confirming properties of continuity in manageable sub-segments of the function's domain.
Thus, neighborhoods serve as a practical tool to harness the predictable nature of continuous functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f\) be a continuous function with domain \(D\) and range contained in \(D\), and let \(a \in D\). Suppose that we define a sequence \(\left(x_{n}\right)\) by the recurrence relation $$ x_{1}=a \text { and } x_{n}=f\left(x_{n-1}\right) \text { for } n>1 $$ and that the sequence converges to \(x_{0}\). Show that \(x_{0}=f\left(x_{0}\right)\). Note that the function \(f\) given by $$ f(x)=\frac{x^{3}+2 x^{2}+1}{8} \quad x \in \mathbb{R} $$ is continuous. Let \(x_{1}=0\) and use \(f\) to define a sequence \(\left(x_{n}\right)\) in the above way. Use your calculator to work out the first few terms of this sequence and hence find, to three decimal places, a root of the equation $$ x^{3}+2 x^{2}-8 x+1=0 $$

Let \(f\) be the function given by $$ f(x)= \begin{cases}\frac{|x|^{2}+x|x|-2}{x-1} & x \neq 1 \\ 3 & x=1\end{cases} $$ Show that for each \(x_{0} \in \mathbb{R}\) the limit \(\lim _{x \rightarrow x_{0}} f(x)\) exists and that for \(x_{0} \neq 1\) the limit equals \(f\left(x_{0}\right)\) itself. Sketch the graph of \(f\). To what number would you change the ' 3 ' in the definition of \(f\) in order to get a function whose graph does not have a 'break"?

Let \(a>0\). Show that for each pair of positive numbers \(x\) and \(y\) we have $$ \log _{a}(x y)=\log _{a} x+\log _{a} y \text { and } \log _{a}\left(\frac{x}{y}\right)=\log _{a} x-\log _{a} y $$

Show that a cubic equation (i.e. one of the form \(a x^{3}+b x^{2}+\) \(c x+d=0\) where \(a \neq 0\) ) has at least one real root.

Suppose that \(f\) and \(g\) are functions with \(g\) continuous at some point \(x_{0}\) and \(f\) continuous at \(g\left(x_{0}\right)\). Show that the composite function \(f \circ g\) is continuous at \(x_{0}\). Show that the function \(h(x)=\left[\log _{2} x\right]\) - the integer part of the \(\log _{2} x\) is continuous at \(x=3\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.