Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 71
Graph the functions in Exercises \(71-74 .\) Then find the extreme values of the function on the interval and say where they occur. $$ f(x)=|x-2|+|x+3|, \quad-5 \leq x \leq 5 $$
Short Answer
Expert verified
Min value is 2 at \(x=-3\); max value is 11 at \(x=5\).
Step by step solution
01
Understanding the Function
The function given is composed of two absolute value expressions: \( f(x) = |x-2| + |x+3| \). This results in a piecewise function where different expressions apply based on the value of \( x \), specifically depending on when the expressions inside the absolute value change sign.
02
Identifying Critical Points
The critical points for the absolute value expressions \(|x-2|\) and \(|x+3|\) occur at \(x = 2\) and \(x = -3\) respectively. These points define where the expressions change their behavior, creating pieces in the piecewise function.
03
Analyzing Each Piece
1. For \(x < -3\), both terms are negative, so \(f(x) = -(x-2) - (x+3) = -2x - 1\).2. For \(-3 \leq x < 2\), the term \(x+3\) becomes positive and \(x-2\) remains negative, so \(f(x) = -(x-2) + (x+3) = x + 5\).3. For \(x \geq 2\), both terms become positive, so \(f(x) = (x-2) + (x+3) = 2x + 1\).
04
Evaluate at Interval and Critical Points
Calculate the function values at the endpoints and critical points: - At \(x = -5\), \(f(-5) = -2(-5) - 1 = 9\).- At \(x = -3\), \(f(-3) = (-3) + 5 = 2\).- At \(x = 2\), \(f(2) = 2(2) + 1 = 5\).- At \(x = 5\), \(f(5) = 2(5) + 1 = 11\).
05
Identify Extreme Values
Compare the values calculated:- The minimum value is \(2\), occurring at \(x = -3\).- The maximum value is \(11\), occurring at \(x = 5\). The extreme values are determined by considering both the endpoints and critical points of the given interval.
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.
Graphing Functions
Graphing functions is a fundamental skill in mathematics that involves plotting a set of points on a coordinate plane to represent a function. When we graph a function like \( f(x) = |x-2| + |x+3| \), it helps us visualize how the function behaves over its domain.
To graph a function accurately, follow these steps:
To graph a function accurately, follow these steps:
- Identify key points such as intercepts, turning points, and critical points.
- Determine the behavior of the function as \( x \) approaches infinity or negative infinity.
- Evaluate the function at several points within its domain to gain insight into its shape.
Piecewise Function
A piecewise function consists of multiple sub-functions, each applying to a certain interval of the domain. The given function \( f(x) = |x-2| + |x+3| \) is an excellent example of a piecewise function because it is defined by different expressions depending on the intervals of \( x \).
The breakdown is as follows:
The breakdown is as follows:
- For \( x < -3 \), the expression is \( -2x - 1 \).
- For \( -3 \leq x < 2 \), the expression is \( x + 5 \).
- For \( x \geq 2 \), the expression is \( 2x + 1 \).
Critical Points
Critical points are vital in understanding a function's graphical behavior because they represent where the derivative is zero or undefined, often indicating local maxima or minima.
In the context of \( f(x) = |x-2| + |x+3| \), the critical points occur at \( x = -3 \) and \( x = 2 \). These points are where the expressions \(|x-2|\) and \(|x+3|\) change signs, affecting the shape and slope of the graph.
To determine critical points:
In the context of \( f(x) = |x-2| + |x+3| \), the critical points occur at \( x = -3 \) and \( x = 2 \). These points are where the expressions \(|x-2|\) and \(|x+3|\) change signs, affecting the shape and slope of the graph.
To determine critical points:
- Delineate the intervals defined by the points of change.
- Examine the sign changes of the derivative within these intervals.
- Evaluate how these points affect the continuity and differentiability of the function.
Absolute Value Function
An absolute value function refers to a function that involves expressions enclosed within absolute value symbols, optimizing the output for non-negative values.
The expression \(|x-2|\) changes to \(x-2\) if \(x\geq 2\), and \(2-x\) if \(x<2\). Similarly, \(|x+3|\) changes to \(x+3\) if \(x\geq -3\) and \(-x-3\) if \(x<-3\).
It's essential to grasp the behavior of absolute value functions as they often involve sudden changes in slope and direction, leading to distinctive V-shaped graphs. Recognizing how:
The expression \(|x-2|\) changes to \(x-2\) if \(x\geq 2\), and \(2-x\) if \(x<2\). Similarly, \(|x+3|\) changes to \(x+3\) if \(x\geq -3\) and \(-x-3\) if \(x<-3\).
It's essential to grasp the behavior of absolute value functions as they often involve sudden changes in slope and direction, leading to distinctive V-shaped graphs. Recognizing how:
- The function handles different sign changes.
- Impact its symmetry and periodic behavior.
- Influence on optimization problems when evaluating distances or differences between two points.
