/*! 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 7 What is the domain of \(f(x)=e^{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 7

What is the domain of \(f(x)=e^{x} / x\) and where is \(f\) continuous?

Short Answer

Expert verified
Answer: The function \(f(x) = \frac{e^x}{x}\) is continuous on the interval \((-\infty, 0) \cup (0, \infty)\).

Step by step solution

01

Identify the Function

The function given is \(f(x) = \frac{e^x}{x}\). This is a rational function where the numerator is an exponential function, \(e^x\), and the denominator is a linear function, \(x\).
02

Find the Domain of the Function

The domain of a function is the set of all possible input values (x-values) for which the function is defined. A rational function is undefined when the denominator is equal to zero. In this case, the denominator is \(x\), so we need to find when \(x\) is equal to zero. So, the function is undefined when \(x=0\). The domain of \(f(x)\) is all real numbers except x = 0. In interval notation, the domain is represented as \((-\infty, 0) \cup (0, \infty)\).
03

Analyze Continuity of the Function

A function is continuous at a point if the value of the function at the point is equal to the limit of the function as x approaches the point. In this case, the exponential function \(e^x\) is continuous for all real numbers, and the linear function \(x\) is also continuous for all real numbers. Since both the numerator and the denominator are continuous for all real numbers, the rational function \(f(x)\) will also be continuous for all real numbers except where the denominator is zero (x=0).
04

State the Function's Continuity

The function \(f(x) = \frac{e^x}{x}\) is continuous on the interval \((-\infty, 0) \cup (0, \infty)\).

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 91影视!

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

Determine the value of the constant \(a\) for which the function $$f(x)=\left\\{\begin{array}{ll} \frac{x^{2}+3 x+2}{x+1} & \text { if } x \neq-1 \\\a & \text { if } x=-1\end{array}\right.$$ is continuous at -1.

Suppose \(f\) is continuous at \(a\) and assume \(f(a)>0 .\) Show that there is a positive number \(\delta>0\) for which \(f(x)>0\) for all values of \(x\) in \((a-\delta, a+\delta) .\) (In other words, \(f\) is positive for all values of \(x\) in the domain of \(f\) and in some interval containing \(a .)\)

Evaluate the following limits, where \(c\) and \(k\) are constants. \(\lim _{h \rightarrow 0} \frac{100}{(10 h-1)^{11}+2}\)

a. Use the identity \(\sin (a+h)=\sin a \cos h+\cos a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \sin x=0\) to prove that \(\lim _{x \rightarrow a} \sin x=\sin a\) thereby establishing that \(\sin x\) is continuous for all \(x\). (Hint: Let \(h=x-a\) so that \(x=a+h\) and note that \(h \rightarrow 0\) as \(x \rightarrow a\).) b. Use the identity \(\cos (a+h)=\cos a \cos h-\sin a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \cos x=1\) to prove that \(\lim _{x \rightarrow a} \cos x=\cos a\).

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0< x-a<\delta$$ Assume fexists for all \(x\) near a with \(x < \) a. We say the limit of \(f(x)\) as \(x\) approaches a from the left of a is \(L\) and write \(\lim _{x \rightarrow a^{-}} f(x)=L,\) if for any \(\varepsilon > 0 \) there exists \(\delta > 0\) such that $$|f(x)-L| < \varepsilon \quad \text { whenever } \quad 0< a-x <\delta$$ Prove that \(\lim _{x \rightarrow 0^{+}} \sqrt{x}=0\).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation
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.