/*! 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 For each of the following prompt... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 7

For each of the following prompts, sketch a graph on the provided axes of a function that has the stated properties. a. \(y=f(x)\) such that - \(f(-2)=2\) and \(\lim _{x \rightarrow-2} f(x)=1\) \- \(f(-1)=3\) and \(\lim _{x \rightarrow-1} f(x)=3\) \- \(f(1)\) is not defined and \(\lim _{x \rightarrow 1} f(x)=0\) \- \(f(2)=1\) and \(\lim _{x \rightarrow 2} f(x)\) does not exist. b. \(y=g(x)\) such that \- \(\text { - } g(-2)=3, g(-1)=-1, g(1)=-2, \text { and } g(2)=3\) \- At \(x=-2,-1,1\) and \(2, g\) has a limit, and its limit equals the value of the function at that point. \- \(g(0)\) is not defined and \(\lim _{x \rightarrow 0} g(x)\) does not exist.

Short Answer

Expert verified
From given points and limits, y=f(x) has discontinuities at -2, 1, and 2 with specific limits; y=g(x) is continuous at -2, -1, 1, and 2 but discontinuous at 0.

Step by step solution

01

Analyze given points and limits of y=f(x)

First, identify the points: 1. At \(x = -2, f(-2) = 2\), but \(\lim _{x \rightarrow -2} f(x) = 1\). 2. At \(x = -1, f(-1) = 3\) and \(\lim _{x \rightarrow -1} f(x) = 3\). 3. At \(x = 1\), \(f(1)\) is not defined, and \(\lim _{x \rightarrow 1} f(x) = 0\). 4. At \(x = 2, f(2) = 1\) and \(\lim _{x \rightarrow 2} f(x)\) does not exist.
02

Sketch y=f(x) for f(-2) and its limit

Mark a point at \(x = -2\) with \(f(-2) = 2\). Also, indicate that as \(x\) approaches \( -2\), the function approaches \(y = 1\) by drawing a point at \(x = -2, y = 1\) and a small open circle around it.
03

Sketch y=f(x) for f(-1) and its limit

At \(x = -1\), draw a point at \(f(-1) = 3\). Since the limit as \(x\) approaches \(-1\) is \(3\), there is no discontinuity here.
04

Sketch y=f(x) for f(1) and its limit

For \(x = 1\), indicate that \(f(1)\) is not defined by leaving a blank space or an open circle around \(x = 1\). However, the limit as \(x\) approaches \(1\) is \(y = 0\), so draw the approaching lines or curve towards \( (1,0) \) but leave \( (1,0) \) open or undefined.
05

Sketch y=f(x) for f(2) and its limit

At \(x = 2\), draw a point at \(f(2) = 1\). Since the limit does not exist, show that approaching from different sides toward \(x = 2\) yields different values. Draw two different functions heading to that \(x\) point.
06

Summary for y=f(x)

Combine all details to complete the sketch of \(y = f(x)\) with discontinuities at \(x = -2\), \(x = 1\), and different limits at \(x = 2\), ensuring the function aligns with given values and limits.
07

Analyze given points and limits of y=g(x)

Identify the points: 1. At \(x = -2, g(-2) = 3\). 2. At \(x = -1, g(-1) = -1\). 3. At \(x = 1, g(1) = -2\). 4. At \(x = 2, g(2) = 3\). At these points, the limit equals the value of the function.
08

Sketch y=g(x) for each point

Plot points at \(x = -2, y = 3\); \(x = -1, y = -1\); \(x = 1, y = -2\); and \(x = 2, y = 3\). Indicate that the limits at these points equal the function values by making smooth connections or dots indicating no discontinuity at these points.
09

Sketch y=g(x) around x=0

For \(x = 0\), indicate that \(g(0)\) is not defined and the limit does not exist. Leave an open space or no connection around \(x = 0\), showing a discontinuity or break in the graph.
10

Summary for y=g(x)

Combine all plotted points and the discontinuity at \(x = 0\) to complete the graph. Ensure it meets all specified conditions.

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Key Concepts

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

limit of a function
In calculus, the limit of a function is the value that the function approaches as the input (or the independent variable) approaches a certain value. Limits are essential for defining concepts like continuity, derivatives, and integrals. The notation for the limit of a function as x approaches a specific value is written as \(\text{lim}_{x \to c} f(x) = L\). Here, if x gets closer to c, then f(x) gets closer to L.
For example, in the given exercise, let's consider the limit for the function y = f(x) at x = -2. The function value is f(-2) = 2, but the limit as x approaches -2 is 1. This tells us that as x gets very close to -2, the nearby values of the function f(x) are near 1, not 2.
Limits are fundamental for understanding more complex calculus topics and play a crucial role in graph sketching.
discontinuous functions
A function is discontinuous where it isn鈥檛 smooth or uninterrupted. Discontinuities are points on the graph where the function is not defined or takes a sudden jump. There are various types of discontinuities, including point discontinuities, jump discontinuities, and infinite discontinuities.
In the provided exercise, y = f(x) has discontinuities at specified points:
  • At x = -2: the function value is 2, but the limit is 1. This jump from 2 to 1 represents a discontinuity.
  • At x = 1: the function is not defined, but the limit as x approaches 1 is 0. This creates a gap in the graph.
  • At x = 2: the function value is 1 but the limit does not exist indicating another type of discontinuity.
Understanding these discontinuities helps in accurately sketching the graph.
function properties
A function's properties refer to the characteristics and behaviors of the function. These include continuity, limits, derivatives, and more. For graph sketching, key properties are:
  • Points and intervals where the function is continuous or discontinuous.
  • Where the function has specific values and how it behaves near those points.

For example, in the exercise鈥檚 y = g(x):
  • At x = -2, g(-2) = 3 and the limit is also 3, indicating no discontinuity.
  • At x = 0, the function is not defined, and the limit does not exist, creating a notable gap in the graph.
Being aware of these properties aids in sketching and interpreting graphs accurately, ensuring all given conditions are met.

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Most popular questions from this chapter

Suppose that the population, \(P\), of China (in billions) can be approximated by the function \(P(t)=1.15(1.014)^{t}\) where \(t\) is the number of years since the start of \(1993 .\) a. According to the model, what was the total change in the population of China between January 1,1993 and January \(1,2000 ?\) What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on January \(1,2000 ?\) Explain and justify, being sure to include proper units on all your answers. b. According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, \(2012 ?\) c. Write an expression involving limits that, if evaluated, would give the exact instantaneous rate of change of the population on today's date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (including units) of the value you have found. d. Find an equation for the tangent line to the function \(y=P(t)\) at the point where the \(t\) -value is given by today's date.

The value, \(V,\) of a particular automobile (in dollars) depends on the number of miles, \(m\), the car has been driven, according to the function \(V=h(m)\). a. Suppose that \(h(40000)=15500\) and \(h(55000)=13200\). What is the average rate of change of \(h\) on the interval \([40000,55000],\) and what are the units on this value? b. In addition to the information given in (a), say that \(h(70000)=11100\). Determine the best possible estimate of \(h^{\prime}(55000)\) and write one sentence to explain the meaning of your result, including units on your answer. c. Which value do you expect to be greater: \(h^{\prime}(30000)\) or \(h^{\prime}(80000)\) ? Why? d. Write a sentence to describe the long-term behavior of the function \(V=h(m)\), plus another sentence to describe the long-term behavior of \(h^{\prime}(m) .\) Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing.

The table below shows the number of calories used per minute as a function of an individual's body weight for three sports: $$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Activity } & 100 \mathrm{lb} & 120 \mathrm{lb} & 150 \mathrm{lb} & 170 \mathrm{lb} & 200 \mathrm{lb} & 220 \mathrm{lb} \\ \hline \text { Walking } & 2.7 & 3.2 & 4 & 4.6 & 5.4 & 5.9 \\ \hline \text { Bicycling } & 5.4 & 6.5 & 8.1 & 9.2 & 10.8 & 11.9 \\ \hline \text { Swimming } & 5.8 & 6.9 & 8.7 & 9.8 & 11.6 & 12.7 \\ \hline \end{array}$$ a) Determine the number of calories that a 200 lb person uses in one half-hour of walking. b) Who uses more calories, a \(120 \mathrm{lb}\) person swimming for one hour, or a \(220 \mathrm{lb}\) person bicycling for a half-hour? c) Does the number of calories of a person swimming increase or decrease as weight increases?

A bungee jumper dives from a tower at time \(t=0\). Her height \(h\) (measured in feet) at time \(t\) (in seconds) is given by the graph in Figure 1.1.4. In this problem, you may base your answers on estimates from the graph or use the fact that the jumper's height function is given by \(s(t)=100 \cos (0.75 t) \cdot e^{-0.2 t}+100\). a. What is the change in vertical position of the bungee jumper between \(t=0\) and \(t=15 ?\) b. Estimate the jumper's average velocity on each of the following time intervals: [0,15] , \([0,2],[1,6],\) and \([8,10] .\) Include units on your answers. c. On what time interval(s) do you think the bungee jumper achieves her greatest average velocity? Why? d. Estimate the jumper's instantaneous velocity at \(t=5\). Show your work and explain your reasoning, and include units on your answer. e. Among the average and instantaneous velocities you computed in earlier questions, which are positive and which are negative? What does negative velocity indicate?

Use linear approximation to approximate \(\sqrt{25.3}\) as follows. Let \(f(x)=\sqrt{x}\). The equation of the tangent line to \(f(x)\) at \(x=25\) can be written in the form \(y=m x+b\). Compute \(m\) and \(b\). \(m=\) \(\square\) \(b=\) \(\square\) Using this find the approximation for \(\sqrt{25.3}\). Answer: \(\square\)
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