/*! 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 53 Describe the interval(s) on whic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Describe the interval(s) on which the function is continuous. $$ f(x)=\frac{x}{x^{2}+1} $$

Short Answer

Expert verified
The function \( f(x) = \frac{x}{x^{2} + 1} \) is continuous on the interval \(-\infty, +\infty \)

Step by step solution

01

Understanding the function

The given function is a rational function and it takes the form \( f(x) = \frac{x}{x^{2} + 1} \), a simple form of a rational function as a ratio of two polynomial functions.
02

Identifying potential discontinuities

Rational functions typically have discontinuities where the denominator is equal to 0. In this case, our denominator is \(x^{2} + 1\). There is no real value of x that can make our denominator equal to 0. Hence, the function is defined for all real numbers.
03

Checking for other types of discontinuities

Other potential discontinuities could arise from some functional transformations, but there is no indication of such in our function. Therefore, our rational function is continuous everywhere in its domain since there are no places where the function goes undefined, or where a discontinuity can occur.
04

Concluding the intervals of continuity

Since there are no discontinuities in our function, we can safely state that our function is continuous everywhere. This can be described as the function being continuous on the interval \(-\infty, +\infty \)

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!

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

Sketch the graph of the function and describe the interval(s) on which the function is continuous. $$ f(x)=\frac{2 x^{2}+x}{x} $$

Find the limit (if it exists). \(\lim _{x \rightarrow 2} f(x),\) where \(f(x)=\left\\{\begin{array}{ll}{4-x,} & {x \neq 2} \\ {0} & {x=2}\end{array}\right.\)

Consumer Awareness The United States Postal Service first class mail rates are \(\$ 0.41\) for the first ounce and \(\$ 0.17\) for each additional ounce or fraction thereof up to 3.5 ounces. A model for the cost \(C\) (in dollars) of a first class mailing that weighs 3.5 ounces or less is given below. $$ C(x)=\left\\{\begin{array}{ll}{0.41,} & {0 \leq x \leq 1} \\ {0.58,} & {1 < x \leq 2} \\ {0.75,} & {2 < x \leq 3} \\ {0.92,} & {3 < x \leq 3.5}\end{array}\right. $$ (a) Use a graphing utility to graph the function and discuss its continuity. At what values is the function not continuous? Explain your reasoning. (b) Find the cost of mailing a 2.5 -ounce letter.

use a graphing utility to graph the function. Then use the Horizontal Line Test to determine whether the function is one-to-one. If it is, find its inverse function. $$ f(x)=2\left(3 x^{2}-\frac{6}{x}\right) $$

The cost (in dollars) of removing \(p \%\) of the pollutants from the water in a small lake is given by \(C=\frac{25,000 p}{100-p}, \quad 0 \leq p<100\) where \(C\) is the cost and \(p\) is the percent of pollutants. (a) Find the cost of removing \(50 \%\) of the pollutants. (b) What percent of the pollutants can be removed for \(\$ 100,000 ?\) (c) Evaluate \(\lim _{p \rightarrow 100^{-}} C .\) Explain your results.

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.