/*! 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 33 Sketch the following curves, ind... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 33

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

Short Answer

Expert verified
The quadratic function \(f(x) = ax^2 + bx + c\) does not have any inflection points since the second derivative is constant and does not change sign.

Step by step solution

01

- Understand the given function

The function given is a quadratic function of the form \[f(x) = ax^2 + bx + c\]where \(a, b,\) and \(c\) are fixed numbers and \(a eq 0\).
02

- Find the first derivative

To find the critical points (relative extrema), first find the first derivative of the function:\[f'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b\]
03

- Find the second derivative

Now, find the second derivative to determine concavity and potential inflection points:\[f''(x) = \frac{d}{dx}(2ax + b) = 2a\]
04

- Analyze the second derivative

The second derivative \(f''(x) = 2a\) is a constant. Since the second derivative does not depend on \(x\), it is either always positive (for \(a > 0\)) or always negative (for \(a < 0\)). This indicates the concavity is consistent across all \(x\).
05

- Determine inflection points

An inflection point occurs where the second derivative changes sign. Since \(f''(x) = 2a\) does not change sign, the function \(f(x)\) does not have any inflection points.
06

- Determine relative extreme points

To find the relative extrema, set the first derivative equal to zero and solve for \(x\):\[2ax + b = 0\]\[x = -\frac{b}{2a}\]This gives the x-coordinate of the vertex of the parabola. This point is either a minimum (if \(a > 0\)) or a maximum (if \(a < 0\)). Evaluate \(f(x)\) at this point to find the y-coordinate.
07

- Sketch the function

Plot the vertex found in the previous step. Since the quadratic function opens upwards (if \(a > 0\)) or downwards (if \(a < 0\)), sketch the parabola accordingly making sure to indicate the vertex.

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.

Quadratic Function
A quadratic function is a polynomial function of degree 2. It can be expressed in the form where 饾憥 鈮� 0.
The graph of a quadratic function is a parabola, which opens upwards if 饾憥 > 0 and downwards if 饾憥 < 0. For example, the function 饾憮(饾懃) = 饾憥饾懃虏 + 饾憦饾懃 + 饾憪 has a symmetrical parabolic shape centered around its vertex.
The coefficients 饾憦 and 饾憪 determine the position and shape of the parabola, but the opening direction is solely determined by 饾憥.
Let's move on to understanding how to differentiate quadratic functions to find their key properties.
First Derivative
The first derivative of a function helps us find the slope at any point and is crucial for locating relative extrema (maxima and minima).
For the quadratic function This gives us a linear equation .
The first derivative .
By setting the first derivative equal to zero and solving for 饾懃, we can find the 饾懃-coordinate of the function's vertex. This point could be a relative maximum or minimum depending on the direction of the parabola.
We can find it by solving the equation . This helps pinpoint the location of the critical points of the quadratic function.
Second Derivative
The second derivative of a function gives us information about the concavity or the curvature of the function graph.
For our quadratic function . This number informs us if the parabola is concave up or concave down across its domain.
Because , the second derivative does not depend on 饾懃 and is a constant. This tells us that the concavity is consistent throughout the graph 鈥� always concave up if 饾憥 > 0 and concave down if 饾憥 < 0.
This also means that there are no inflection points in a quadratic function, as inflection points are only present where the concavity changes.
Relative Extrema
Relative extrema refer to the highest or lowest points within a certain interval on the function graph, known as maximum and minimum points.
For a quadratic function , there is one relative extremum.
We find it by solving the equation .
The point helps us determine the type of extremum.
When you substitute , you get the coordinates of the vertex. If 饾憥 > 0, this point is a minimum because the parabola opens upwards. If 饾憥 < 0, this point is a maximum because the parabola opens downwards.
Concavity
Concavity describes whether a graph bends upwards or downwards.
For a quadratic function like , depending on whether 饾憥 is positive or negative, the parabola can either open upwards (concave up) or downwards (concave down).
When the second derivative 饾憮''(饾懃) is positive (饾憥 > 0), the graph is concave up like a U.
When the second derivative 饾憮''(饾懃) is negative (饾憥 < 0), the graph is concave down like a n.
This property is consistent across all points on the graph of a quadratic function.

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

Each of the graphs of the functions has one relative maximum and one relative minimum point. Plot these two points and check the concavity there. Using only this information, sketch the graph. $$f(x)=\frac{1}{3} x^{3}+2 x^{2}-5 x+\frac{8}{3}$$

Some years ago, it was estimated that the demand for steel approximately satisfied the equation \(p=256-50 x\) and the total cost of producing \(x\) units of steel was \(a(x)=182+56 x\). (The quantity \(x\) was measured in millions of tons and the price and total cost were measured in millions of dollars.) Determine the level of production and the corresponding price that maximize the profits.

Each of the graphs of the functions has one relative maximum and one relative minimum point. Plot these two points and check the concavity there. Using only this information, sketch the graph. $$f(x)=2 x^{3}-15 x^{2}+36 x-24$$

The graph of each function has one relative extreme point. Find it (giving both \(x\) - and \(y\) -coordinates) and determine if it is a relative maximum or a relative minimum point. Do not include a sketch of the graph of the function. $$f(x)=5 x^{2}+x-3$$

A farmer has \(\$ 1500\) available to build an E-shaped fence along a straight river so as to create two identical rectangular pastures. (See Fig. 13.) The materials for the side parallel to the river cost \(\$ 6\) per foot, and the materials for the three sections perpendicular to the river cost \(\$ 5\) per foot. Find the dimensions for which the total area is as large as possible.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

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.