Chapter 5: Problem 62
In Exercises \(55-62,\) graph the function and find its average value over the given interval. $$ h(x)=-|x| \quad \text { on } \quad \text { a. }[-1,0], \text { b. }[0,1], \text { and } c .[-1,1] $$
Short Answer
Expert verified
The average value of \( h(x) = -|x| \) over each interval is \(-\frac{1}{2}\).
Step by step solution
01
Understand the Function
The function given is \( h(x) = -|x| \). This function represents a V-shaped graph pointing downward, symmetric about the y-axis, because the absolute value \(|x|\) is being negated.
02
Graph the Function
To graph \( h(x) = -|x| \), plot the points for some values of \(x\):\( x = -2, h(x) = -2 \), \( x = -1, h(x) = -1 \), \( x = 0, h(x) = 0 \), \( x = 1, h(x) = -1 \), \( x = 2, h(x) = -2 \). Connect the points to form a V shape that opens downward with the vertex at (0,0).
03
Find the Average Value Formula
The formula to find the average value of a function \( f(x) \) over an interval \([a, b]\) is \( \frac{1}{b-a} \int_a^b f(x) \, dx \).
04
Average Value Over \([-1, 0]\)
For the interval \([-1, 0]\), calculate \( \frac{1}{0 - (-1)} \int_{-1}^{0} -|x| \, dx = \int_{-1}^{0} x \, dx \) because \(-|x| = x\) for \(x < 0\). The integral \( \int_{-1}^{0} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{0} = 0 - \frac{(-1)^2}{2} = -\frac{1}{2} \). Hence, the average value is \(-\frac{1}{2}\).
05
Average Value Over \([0, 1]\)
For the interval \([0, 1]\), use \( \frac{1}{1 - 0} \int_{0}^{1} -|x| \, dx = \int_{0}^{1} -x \, dx \). The integral \( \int_{0}^{1} -x \, dx = \left[ -\frac{x^2}{2} \right]_{0}^{1} = -\frac{1}{2} - 0 = -\frac{1}{2} \). Thus, the average value is \(-\frac{1}{2}\).
06
Average Value Over \([-1, 1]\)
For the interval \([-1, 1]\), split the integral at zero: \( \int_{-1}^{1} -|x| \, dx = \int_{-1}^{0} x \, dx + \int_{0}^{1} -x \, dx \). We have already calculated these as \(-\frac{1}{2}\) each. The sum is \(-\frac{1}{2} - \frac{1}{2} = -1\). The average value is \( \frac{1}{2 \times 1}(-1) = -\frac{1}{2} \).
07
Conclusion
The average value of \( h(x) = -|x| \) over each interval, \([-1,0]\), \([0,1]\), and \([-1,1]\), is \(-\frac{1}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Functions
Graphing functions is like drawing a picture of how a function behaves across different values of its domain. For the function \( h(x) = -|x| \), we are looking at a graph that reflects symmetry and a specific structure due to the absolute value involved.
When you graph an absolute value function, it typically forms a 'V' shape. This happens because absolute value measures distance, effectively transforming all inputs into non-negative outputs. However, in this case, we've added a negative sign in front of the absolute value which impacts the orientation of the graph.
When you graph an absolute value function, it typically forms a 'V' shape. This happens because absolute value measures distance, effectively transforming all inputs into non-negative outputs. However, in this case, we've added a negative sign in front of the absolute value which impacts the orientation of the graph.
- The graph of \( h(x) = -|x| \) resembles an upside-down 'V', or more accurately, a caret pointing downward. This is due to the negative sign before the absolute value.
- Its vertex is at the origin, \((0,0)\), where the graph changes direction.
- The line symmetry means for every point \((x, y)\), there is a corresponding point \((-x, y)\).
Integrals
Integrals play a crucial role in understanding how a function behaves over an interval. They help us calculate areas under curves and are vital for finding the average value of a function over a specific section of its domain.
To find the average value of a function, like \( h(x) = -|x| \), over an interval \([a, b]\), use the formula:
\[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
Here's why this works:
To find the average value of a function, like \( h(x) = -|x| \), over an interval \([a, b]\), use the formula:
\[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
Here's why this works:
- The integral calculates the total area or 'accumulated quantity' under the curve between \(a\) and \(b\).
- The formula divides this total area by the length \(b-a\) of the interval to find an average height of the function in this interval.
Absolute Value Function
The absolute value function is a mathematical operation that returns a number's non-negative value, regardless of its original sign. When you apply negative scaling to an absolute value function like \( h(x) = -|x| \), it turns the functional graph on its head, creating a new visual representation.
Key aspects of absolute value functions include:
Key aspects of absolute value functions include:
- They produce a 'V' shaped graph when plotted, centered at zero, illustrating their symmetry about the vertical axis.
- When negated, such as in \( h(x) = -|x| \), the graph is flipped downward.
- This function can be split into piecewise components. For instance, \(-|x|\) outputs \(-x\) when \(x\) is positive, and \(x\) when \(x\) is negative, influencing how integrals are computed.