/*! 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 Use the given information to mak... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 14

Use the given information to make a good sketch of the function \(f(x)\) near \(x=3\). $$f(3)=-2, f^{\prime}(3)=0, f^{\prime \prime}(3)=1$$

Short Answer

Expert verified
Sketch a curve with a local minimum at \((3, -2)\), a horizontal tangent at \(x = 3\), and the function concave up around this point.

Step by step solution

01

Interpret the Given Information

The point of interest is at \(x = 3\). The function value at this point is \(f(3) = -2\). The first derivative at this point is \(f^{\textprime}(3) = 0\), which indicates a potential local maximum, minimum, or inflection point. The second derivative at this point is \(f^{\textprime \textprime}(3) = 1\), which is positive, indicating that the function is concave up at \(x = 3\).
02

Plot the Key Point

Start by plotting the point \((3, -2)\) on the graph. This is the point where the function value is provided.
03

Analyze the First Derivative

Since \(f^{\textprime}(3) = 0\), the slope of the tangent to the curve at \(x = 3\) is horizontal. Draw a horizontal tangent line at the point \((3, -2)\).
04

Conclude the Behavior Using the Second Derivative

Given that \(f^{\textprime \textprime}(3) = 1\), the function is concave up at \(x = 3\). This suggests that \((3, -2)\) is a local minimum point.
05

Sketch the Function near \(x=3\)

Using the information above, sketch a curve near \(x = 3\) such that it has a local minimum at the point \((3, -2)\). Ensure the curve flattens out into a horizontal tangent at this point before continuing to curve upwards as it moves away from \(x = 3\).

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.

Local Minimum
A local minimum of a function is a point where the function has a lower value than at any nearby points. For the function given in the exercise, we see that the value of the first derivative at \(x = 3\) is zero, i.e., \(f^{\textprime}(3) = 0\). This tells us that \(x = 3\) is a critical point. To determine whether this critical point is a local minimum, maximum, or a point of inflection, we check the second derivative. In this case, the second derivative at \(x = 3\) is positive, \(f^{\textprime \textprime}(3) = 1\), which indicates the function is concave up, confirming that the point \((3, -2)\) is indeed a local minimum.
  • Critical point: \(f^{\textprime}(3) = 0\)
  • Concave up: \(f^{\textprime \textprime}(3) > 0\)
  • Local minimum: Point \((3, -2)\)
First Derivative
The first derivative of a function, denoted as \(f^{\textprime}(x)\), represents the slope of the tangent line to the curve at any given point. In the exercise, \(f^{\textprime}(3) = 0\) implies that the slope of the tangent line at \(x = 3\) is zero. This means the tangent at \((3, -2)\) is horizontal. Horizontal tangents occur at local maxima, minima, or inflection points.

To sketch the function correctly:
  • Evaluate \(f^{\textprime}(x)\) to find critical points
  • In this problem, \(f^{\textprime}(3) = 0\) indicates a horizontal tangent at \(x = 3\)
Second Derivative
The second derivative, \(f^{\textprime \textprime}(x)\), provides information about the concavity of the function. In this exercise, it's given that \(f^{\textprime \textprime}(3) = 1\). A positive second derivative signifies that the function is concave up at the point \(x = 3\).

Key points about second derivatives:
  • If \(f^{\textprime \textprime}(x) > 0\), the function is concave up at that point
  • If \(f^{\textprime \textprime}(x) < 0\), the function is concave down at that point
  • Used to identify the nature of critical points detected by the first derivative
Concavity
Concavity describes the direction the function curves. When a function is concave up, it looks similar to a 'U' shape; when concave down, it forms an 'n' shape. In this exercise, because \(f^{\textprime \textprime}(3) = 1\) is positive, the function is concave up at \(x = 3\). This information confirms the presence of a local minimum at \((3, -2)\).

Understanding concavity:
  • A function concave up \rightarrow \(f^{\textprime \textprime}(x) > 0\)
  • A function concave down \rightarrow \(f^{\textprime \textprime}(x) < 0\)
  • Determines shaping of the curve near critical points

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 following curves, indicating all relative extreme points and inflection points. Let \(a, b, c, d\) be fixed numbers with \(a \neq 0,\) and let \(f(x)=a x^{3}+b x^{2}+c x+d .\) Is it possible for the graph of f(x) to have more than one inflection point? Explain your answer.

Each of the graphs of the functions has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. [Recall that if \(f(x)=a x^{2}+b x+c,\) then \(f(x)\) has a relative minimum point when \(a>0\) and a relative maximum point when \(a<0.1\) $$f(x)=1+6 x-x^{2}$$

Sketch the graph of a function having the given properties. Relative maximum points at \(x=1\) and \(x=5 ;\) relative minimum point at \(x=3 ;\) inflection points at \(x=2\) and \(x=4\)

Each of the graphs of the functions has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. [Recall that if \(f(x)=a x^{2}+b x+c,\) then \(f(x)\) has a relative minimum point when \(a>0\) and a relative maximum point when \(a<0.1\) $$f(x)=-x^{2}+2 x-5$$

Each of the graphs of the functions has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. [Recall that if \(f(x)=a x^{2}+b x+c,\) then \(f(x)\) has a relative minimum point when \(a>0\) and a relative maximum point when \(a<0.1\) $$f(x)=\frac{1}{2} x^{2}+\frac{1}{2}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

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.