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Magazine Chapter 4: Problem 79
Graph the functions in Exercises \(77-80 .\) Then find the extreme values
of the function on the interval and say where they occur.
$$
h(x)=|x+2|-|x-3|, \quad-\infty
Short Answer
Expert verified
Minimum of -5 at x = -2; maximum of 5 at x = 3.
Step by step solution
01
Understand the Function
The given function is \( h(x) = |x+2| - |x-3| \). This is a piecewise function composed of two absolute value functions. Each absolute value function will change its behavior at its respective critical points which are \(-2\) for \(|x+2|\) and \(3\) for \(|x-3|\).
02
Define Intervals Based on Critical Points
The behavior of the absolute value function changes at \(x = -2\) and \(x = 3\). Therefore, the entire number line is divided into three intervals: \((-fty, -2)\), \([-2, 3)\), and \([3, fty)\). We will consider these intervals separately to determine the piecewise form of the function.
03
Evaluate The Function on Each Interval
**Interval 1: \(x < -2\)**- Here, both \(|x+2|\) and \(|x-3|\) are evaluated as their negative expressions: \(-x-2\) and \(-x+3\).- Therefore, \(h(x) = (-x-2) - (-x+3) = -2 - 3 = -5\).**Interval 2: \(-2 \leq x < 3\)**- Here, \(|x+2|\) is positive so it evaluates as \(x+2\), and \(|x-3|\) evaluates as \(-x+3\).- Therefore, \(h(x) = (x+2) - (-x+3) = x + 2 + x - 3 = 2x - 1\).**Interval 3: \(x \geq 3\)**- Here, both \(|x+2|\) and \(|x-3|\) are positive, so \(h(x) = (x+2) - (x-3) = 5\).
04
Find Extreme Values
To find the extreme values, evaluate \(h(x)\) at the critical points and analyze each interval:1. **Critical Point at \(x = -2\):** - \(h(-2) = |0| - |5| = 0 - 5 = -5\).2. **Critical Point at \(x = 3\):** - \(h(3) = |5| - |0| = 5 - 0 = 5\).Given the intervals, - In \((-fty, -2)\), \( h(x) = -5 \), which would be a minimum as we approach \(x = -2\).- In \([3, fty)\), \(h(x) = 5\) which remains constant, hence maximum at \(x = 3\).
05
Identify and State Extreme Values
The extreme values of the function are:- A minimum of \(-5\) occurring at \(x = -2\).- A maximum of \(5\) occurring at \(x = 3\).This analysis shows where the extreme values occur as per the intervals chosen.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Functions
Piecewise functions are mathematical functions defined by multiple sub-functions, each applying to a specific interval in the domain. In our example with the function \( h(x) = |x+2| - |x-3| \), the function behaves differently based on the values of \( x \).
These behaviors depend on the critical points at \( x = -2 \) and \( x = 3 \), which split the number line into distinct intervals:
These behaviors depend on the critical points at \( x = -2 \) and \( x = 3 \), which split the number line into distinct intervals:
- For \( x < -2 \), both expressions \( |x+2| \) and \( |x-3| \) result in their negative forms, leading to a constant result, \( h(x) = -5 \).
- For \( -2 \leq x < 3 \), the function becomes \( h(x) = 2x - 1 \), a linear function on this interval.
- For \( x \geq 3 \), both absolute functions are positive, simplifying \( h(x) \) to a constant value, \( h(x) = 5 \).
Absolute Value Functions
Absolute value functions are special because they describe the distance from zero, regardless of direction. This plays a key role in the behavior of piecewise functions.
In the exercise function \( h(x) = |x+2| - |x-3| \), each component shifts the graph horizontally:
In the exercise function \( h(x) = |x+2| - |x-3| \), each component shifts the graph horizontally:
- The function \( |x+2| \) translates the standard absolute value graph two units to the left, creating a vertex at \( x = -2 \).
- Similarly, \( |x-3| \) shifts the graph three units to the right, creating a vertex at \( x = 3 \).
- Negative behavior prevails on \( x < -2 \), forcing the negative forms \( -x-2 \) and \(-x+3 \) to be used.
- For \( -2 \leq x < 3 \), the function transitions because \( |x+2| \) stabilizes as positive while \( |x-3| \) remains negative.
- Finally, for \( x \geq 3 \), both values are positive, simplifying to constants.
Critical Points in Calculus
Critical points in a function occur where its derivative is zero or undefined, often signaling potential extreme values such as minima or maxima.
For our function, \( h(x) = |x+2| - |x-3| \), the critical points are identified by settings \( x = -2 \) and \( x = 3 \). These values are where the behavior of \( h(x) \) changes dramatically.
Let's break the analysis process:
For our function, \( h(x) = |x+2| - |x-3| \), the critical points are identified by settings \( x = -2 \) and \( x = 3 \). These values are where the behavior of \( h(x) \) changes dramatically.
Let's break the analysis process:
- Calculate the derivative of each segment within the defined piecewise intervals.
- Assess the behavior at the critical points. The derivative is used for detecting where extrema might occur due to changes in direction or slope.
- At \( x = -2 \), the value is computed as \( h(-2) = -5 \), showing a minimum because the function transitions from \(-5\) to a linear form immediately after.
- At \( x = 3 \), a value of \( h(3) = 5 \) is calculated, showing a maximum sustained across later intervals.
