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Magazine Chapter 4: Problem 89
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 is -5 at x = -2; Maximum is 5 at x = 3.
Step by step solution
01
Understand the Absolute Value Function
The function given is composed of two absolute values: \(h(x) = |x+2| - |x-3|\). The absolute value function \(|a|\) equals \(a\) when \(a \geq 0\) and \(-a\) when \(a < 0\). Identify the points where each absolute value expression changes its behavior: \(x+2=0\) gives \(x=-2\), and \(x-3=0\) gives \(x=3\). These points split the number line into three intervals: \((-\infty, -2)\), \((-2, 3)\), and \((3, \infty)\).
02
Evaluate the Function on Each Interval
For each interval, determine the expression for \(h(x)\):- For \(x < -2\), both \((x+2) < 0\) and \((x-3) < 0\), so \(|x+2| = -(x+2)\) and \(|x-3| = -(x-3)\). Therefore: \[h(x) = -(x+2) + (x-3) = -2x - 5\].- For \(-2 \leq x < 3\), \((x+2) \geq 0\) and \((x-3) < 0\), so \(|x+2| = x+2\) and \(|x-3| = -(x-3)\). Therefore:\[h(x) = (x+2) + (x-3) = 2x - 1\].- For \(x \geq 3\), both \((x+2) \geq 0\) and \((x-3) \geq 0\), so \(|x+2| = x+2\) and \(|x-3| = x-3\). Therefore:\[h(x) = (x+2) - (x-3) = 5\].
03
Find Critical Points
Critical points occur at values of \(x\) where the derivative is zero or undefined or at boundary points of sub-domains.- The intersection points already considered are \(x = -2\) and \(x = 3\). We found segment expressions not needing derivatives since they are linear pieces within each segment.- Calculate \(h(-2)\) and \(h(3)\): - \(h(-2) = |0| - |-5| = 0 - 5 = -5\). - \(h(3) = |5| - |0| = 5\).
04
Determine Extreme Values
By comparing values on each interval:- For \(x < -2\), \(h(x) = -2x - 5\) which decreases further without bound as \(x\) decreases further, but at \(x = -2\), \(h(x) = -5\).- For \(-2 \leq x < 3\), \(h(x) = 2x - 1\) (a linearly increasing function) that takes its minimum value at \(x = -2\), so \(h(x) = -5\).- For \(x \geq 3\), \(h(x) = 5\) is constant, so its maximum throughout entire function is \(x = 3\), \(h(3) = 5\).
05
Conclude Extreme Values
The extreme values occur at \(x = -2\) with \(h(x) = -5\), which is the minimum value in the specific intervals and at \(x = 3\) with \(h(x) = 5\), which is the maximum value for the entire domain. This gives: - Minimum value is \(-5\) at \(x = -2\). - Maximum value is \(5\) at \(x = 3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Function
The absolute value function is a mathematical expression that transforms any real number into its non-negative counterpart. Formally, for any real number \(a\), the absolute value is expressed as \(|a|\). This is equal to \(a\) if \(a \geq 0\) and \(-a\) if \(a < 0\).
- \(|x+2|\): This will be \(x+2\) when \(x \geq -2\), and \(-(x+2)\) when \(x < -2\).
- \(|x-3|\): It is \(x-3\) for \(x \geq 3\), and \(-(x-3)\) when \(x < 3\).
Critical Points
Critical points are vital for finding where a function's behavior might change. In calculus, critical points include places where the function's derivative is zero or undefined. Additionally, they can occur at boundaries or endpoints of distinct sections in a piecewise function. For the function \(h(x) = |x+2| - |x-3|\):
- The identified transition points \(x = -2\) and \(x = 3\) are critical, as they define the edges of the intervals that determine different linear expressions for \(h(x)\).
- To assess critical points, examining the function's behavior on either side of these transition points is essential, as these points might indicate where a potential maximum or minimum value could occur.
Extreme Values
Extreme values of a function are the highest or lowest points over its domain or within a specific interval. They can either be maximum or minimum values. Calculating extreme values involves evaluating function outputs at critical points and comparing them to values inside intervals:For the function \(h(x) = |x+2| - |x-3|\) defined over the entire real number line:
- Minimum Value: Occurs at \(x = -2\) with \(h(-2) = -5\).
- Maximum Value: Achieved at \(x = 3\) where \(h(3) = 5\).
- Within the interval \(-2 \leq x < 3\), \(h(x) = 2x - 1\) is increasing linearly, reaching its lowest at \(x = -2\) and highest just before \(x = 3\).
Derivatives
In calculus, derivatives represent the rate of change or the slope of a function at any point. They are instrumental in identifying the behavior of complex mathematical expressions such as absolute value functions:For \(h(x) = |x+2| - |x-3|\):
- Derivatives at linear sections: Using the derivative concept helps when typical derivatives within linear segments behave predictably, maintaining or shifting slopes based on the sign change of absolute functions.
- On \(x < -2\): The function \(h(x) = -2x-5\) has a constant negative slope, indicating a consistent downward trend.
- Within \(-2 \leq x < 3\): The function is \(h(x) = 2x - 1\) with a positive slope, showing an upward trend.
- Beyond \(x \geq 3\): \(h(x) = 5\) is constant, signifying zero slopes.
