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Chapter 5: Problem 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
a. \(-\frac{1}{2}\), b. \(-\frac{1}{2}\), c. 0

Step by step solution

01

Understanding the Function

The function given is \( h(x) = -|x| \). This is the negative of the absolute value function. The graph of this function is a V-shaped curve opening downwards.
02

Graph the Function

To graph \( h(x) = -|x| \), note that it takes the values \( h(x) = -x \) when \( x \geq 0 \) and \( h(x) = x \) when \( x < 0 \). This creates a straight line with a downward slope on both sides of the y-axis, intersecting at the origin (0,0).
03

Step 3a: Find the Average Value on \([-1,0]\)

Calculate the average value of the function on \([-1,0]\) using the formula for the average value of a function: \( \frac{1}{b-a} \int_a^b h(x) \, dx \). Here, \( a = -1 \) and \( b = 0 \).\[-\int_{-1}^{0} x \, dx = \left[ \frac{-x^2}{2} \right]_{-1}^{0} = \frac{0^2}{2} - \frac{-(-1)^2}{2} = \frac{1}{2}\]Thus, the average value is \(-\frac{1}{1} \times \frac{1}{2} = -\frac{1}{2}\).
04

Step 3b: Find the Average Value on \([0,1]\)

Calculate the average value of the function on \([0,1]\): \[-\int_{0}^{1} x \, dx = \left[ \frac{-x^2}{2} \right]_{0}^{1} = \frac{-1^2}{2} - \frac{0^2}{2} = -\frac{1}{2}\]Thus, the average value is \(-\frac{1}{1} \times -\frac{1}{2} = -\frac{1}{2}\).
05

Step 3c: Find the Average Value on \([-1,1]\)

Calculate the average value of the function on \([-1,1]\):\[\frac{1}{2} \int_{-1}^{1} h(x) \, dx = \frac{1}{2} \left(\int_{-1}^{0} x \, dx + \int_{0}^{1} -x \, dx\right)\]We have already found that \(-\int_{-1}^{0} x \, dx = \frac{1}{2}\) and \(-\int_{0}^{1} x \, dx = -\frac{1}{2}\). Thus:\[\frac{1}{2} \left( \frac{1}{2} - \frac{1}{2} \right) = 0\]The average value of the function on \([-1,1]\) is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphing Functions
When graphing a function like \( h(x) = -|x| \), understanding its behavior is crucial. This function flips the standard absolute value function, \( |x| \), upside down. Graphically, this creates a V-shape opening downwards.
This specific function results in straight lines that slope downwards from the origin.
  • For \( x \geq 0 \), the function behaves as \( h(x) = -x \). This is a straight line with a negative slope.
  • For \( x < 0 \), it acts like \( h(x) = x \) but also extended downward. This results in symmetry around the y-axis.
These characteristics mean the graph is symmetrical and points downward from the vertex at the origin (0,0). The symmetry and slope are key features to correctly graph \( h(x) = -|x| \).
Average Value of a Function
The average value of a function gives us insight into the "typical" value it takes over a given interval. Mathematically, it's calculated with the formula:\[ \text{Average value of } f(x) = \frac{1}{b-a} \int_a^b f(x) \, dx \]This means we integrate the function over the interval from \( a \) to \( b \) and then divide by the length of the interval.
  • For \([-1, 0]\), calculate: \[\frac{1}{0-(-1)} \int_{-1}^{0} x \, dx = -\frac{1}{2} \]
  • For \([0, 1]\), the computation is: \[\frac{1}{1-0} \int_{0}^{1} -x \, dx = -\frac{1}{2} \]
  • For \([-1, 1]\), find: \[\frac{1}{2-(-1)} \int_{-1}^{1} h(x) \, dx = 0 \]
The average value indicates balanced positive and negative values over each respective interval, with \([-1, 1]\) showing complete balance between sections.
Definite Integrals
Definite integrals are vital in calculating areas under curves, which translates directly into the average value of a function in calculus. Specifically, the definite integral of a function from \( a \) to \( b \) is represented as:\[ \int_a^b f(x) \, dx \]This operation gives the net area between the curve and the x-axis over the specified interval.
  • For \( h(x) = -x \) on \([0, 1]\), \ the definite integral computes the area beneath the line, resulting in a negative since it's below the x-axis.
  • Conversely, \( h(x) = x \) on \([-1, 0]\) scores positive due to its position above the x-axis but is negated by the negative sign in \(-|x|\).
  • Over the entirety of \([-1, 1]\), the positive and negative areas effectively cancel each other out, leading to a net integral result of zero.
This balance between positive and negative sections reflects the symmetry in the graph of \( h(x) = -|x| \), inherently representing the function's average value across its domain.

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