/*! 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 34 In each part, identify the domai... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 34

In each part, identify the domain and range of the function, and then sketch the graph of the function without using a graphing utility. $$ \text { (a) } f(x)=1-e^{-x+1} \quad \text { (b) } g(x)=3 \ln \sqrt[3]{x-1} $$

Short Answer

Expert verified
(a) Domain: \((-\infty, \infty)\), Range: \((1, \infty)\). (b) Domain: \((1, \infty)\), Range: \((-\infty, \infty)\).

Step by step solution

01

Identify the Domain of f(x)

For the function \( f(x) = 1 - e^{-x+1} \), the expression inside the exponential function, \( -x+1 \), can take any real number value. Thus, the domain of \( f(x) \) is all real numbers, denoted as \( (-\infty, \infty) \).
02

Identify the Range of f(x)

The exponential function \( e^u \) is always positive and approaches 0 as \( u \to -\infty \) and approaches \( +\infty \) as \( u \to +\infty \). Therefore, \( -e^{-x+1} \) will range from \(-\infty\) to 0. Hence, \( f(x) = 1 - e^{-x+1} \) ranges from 1 (when \( -e^{-x+1} = 0 \)) to \(-\infty\) (as \(-e^{-x+1}\) approaches large negative values), so the range is \((1, \infty)\).
03

Sketch the Graph of f(x)

Start by plotting the horizontal asymptote at \( y = 1 \). As \( x \to \infty \), the term \(-e^{-x+1} \to 0\), thus \( f(x) \to 1 \). Conversely, as \( x \to -\infty \), \( -e^{-x+1} \to -\infty \)', making \( f(x) \to \infty \). Sketch a curve starting above the y-axis, approaching the asymptote \( y = 1 \) from above and rising sharply as you move left.
04

Identify the Domain of g(x)

For \( g(x) = 3 \ln \sqrt[3]{x-1} \), the expression inside the logarithm, \( \sqrt[3]{x-1} \), must be positive, implying \( x-1 > 0 \). Thus, the domain is \( x > 1 \), or \( (1, \infty) \).
05

Identify the Range of g(x)

The cube root and logarithmic functions can produce any real number, since \( \ln(u) \) is defined for all positive \( u \). Therefore, \( g(x) \) can take any real value. The range is \((-\infty, \infty)\).
06

Sketch the Graph of g(x)

Since the logarithmic function increases without bound, sketch the graph starting at \( x = 1 \) from negative infinity (the vertical asymptote \( x = 1 \)) and moving upwards slowly as x increases, reflecting the natural increase and curve of the logarithmic function.

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!

Key Concepts

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

Exponential Function
An exponential function is one where a constant base is raised to a variable exponent. In the function \( f(x) = 1 - e^{-x+1} \), the base \( e \) is the Euler's number, approximately equal to 2.718. The expression \( -x+1 \) in the exponent can vary across all real numbers, which defines the domain of our function as
  • \((-\infty, \infty)\)

Understanding the range of an exponential function is crucial, too. The term \( -e^{-x+1} \) tends to 0 as \( x \) gets larger and negative when \( x \) becomes very negative. Thus, the function \( f(x) \) ranges from 1 to infinity,
  • \((1, \infty)\)

This knowledge helps when sketching the graph, where you plot the horizontal asymptote at \( y = 1 \) and visualize the graph rising without bound.
Logarithmic Function
A logarithmic function helps in reversing the effects of exponential functions. For \( g(x) = 3 \ln \sqrt[3]{x-1} \), understanding the behavior of the logarithm is key. Here, the cube root \( \sqrt[3]{x-1} \) must be positive.
  • Domain: \(x > 1\)
  • Domain Interval: \((1, \infty)\)

The range of \( g(x) \) is unrestricted among real numbers since both cubic root and logarithm can cover any value, hence
  • Range: \((-\infty, \infty)\)

Graphing logarithmic functions begins by identifying any vertical asymptotes; here it lies at \( x = 1 \). Afterwards, the curve increases slowly from negative infinity.
Graph Sketching
Graph sketching is an art for visualizing mathematical functions. For \( f(x) = 1 - e^{-x+1} \), begin with its horizontal asymptote at \( y = 1 \). As \( x \) moves to positive values, \( f(x) \) nears the asymptote from above and rises infinitely as \( x \) becomes negatively large.
For plotting \( g(x) = 3 \ln \sqrt[3]{x-1} \), remember its vertical asymptote at \( x = 1 \). The graph rises slowly to the right, getting increasingly positive as \( x \) grows. This slow ascent is typical for logarithmic functions.
Both functions need careful attention to their asymptotes, ensuring the sketch accurately depicts each's behavior. Use consistent scales on the axes to clarify changes in each direction. When sketching, consistent practice results in more precise and intuitive graphs. But remember, start simple and add complexity gradually.

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

The formula \(F=\frac{9}{5} C+32,\) where \(C \geq-273.15\) expresses the Fahrenheit temperature \(F\) as a function of the Celsius temperature \(C\). (a) Find a formula for the inverse function. the inverse function.

A variable \(y\) is said to be inversely proportional to the square of a variable \(x\) if \(y\) is related to \(x\) by an equation of the form \(y=k / x^{2},\) where \(k\) is a nonzero constant, called the constant of proportionality. This terminology is used in these exercises. According to Coulomb's \(\operatorname{law},\) the force \(F\) of attraction between positive and negative point charges is inversely proportional to the square of the distance \(x\) between them. (a) Assuming that the force of attraction between two point charges is 0.0005 newton when the distance between them is 0.3 meter, find the constant of proportionality (with proper units). (b) Find the force of attraction between the point charges when they are 3 meters apart. (c) Make a graph of force versus distance for the two charges. (d) What happens to the force as the particles get closer and closer together? What happens as they get farther and farther apart?

Find a formula for \(f^{-1}(x)\) $$ f(x)=\sqrt[5]{4 x+2} $$

Use a calculating utility to approximate the expression. Round your answer to four decimal places. $$ \begin{array}{lll}{\text { (a) } \sqrt[5]{24}} & {\text { (b) } \sqrt[8]{0.6}}\end{array} $$

As discussed in this section, Boyle's law states that at a constant temperature the pressure \(P\) exerted by a gas is related to the volume \(V\) by the equation \(P V=k\) (a) Find the appropriate units for the constant \(k\) if pressure (which is force per unit area) is in newtons per square meter (N/m') and volume is in cubic meters (m'). (b) Find \(k\) if the gas exerts a pressure of \(20,000 \mathrm{N} / \mathrm{m}^{2}\) when (c) Make a table that shows the pressures for volumes of \(0.25,0.5,1.0,1.5,\) and 2.0 liters. (d) Make a graph of \(P\) versus \(V\).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

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.