Problem 8
Question
Graph the functions in Exercises \(7-18 .\) What symmetries, if any, do the graphs have? Specify the intervals over which the function is increasing and the intervals where it is decreasing. $$ y=-\frac{1}{x^{2}} $$
Step-by-Step Solution
Verified Answer
The graph is symmetric about the y-axis, decreasing on \((-
fty, 0)\) and \((0,
fty)\).
1Step 1: Understand the Function
The function given is \( y = -\frac{1}{x^2} \). This is a rational function where \( x^2 \) appears in the denominator with a negative sign in front of it. It's important to note that the function is undefined at \( x = 0 \).
2Step 2: Analyze Symmetry
A function is symmetric with respect to the y-axis if \( f(-x) = f(x) \) for all \( x \). Substitute \( -x \) into the function: \( y = -\frac{1}{(-x)^2} = -\frac{1}{x^2} \). We observe that \( f(-x) = f(x) \), indicating symmetry with respect to the y-axis.
3Step 3: Determine Intervals of Increase and Decrease
The function \( y = -\frac{1}{x^2} \) decreases as \( x \) moves away from zero in both the positive and negative directions because the value becomes more negative as \( |x| \) increases. To verify, calculate the derivative: \( y' = \frac{2}{x^3} \). The derivative is positive when \( x < 0 \) and negative when \( x > 0 \), indicating that the function decreases over \((-fty, 0)\) and \((0, fty)\).
4Step 4: Graph the Function
Graphing \( y = -\frac{1}{x^2} \), we see vertical asymptotes at \( x = 0 \) since the function shoots to negative infinity around \( x = 0 \). The graph is a downward opening curve in both \((-fty, 0)\) and \((0, fty)\), confirming that the function decreases on these intervals.
Key Concepts
Function SymmetryIncreasing and Decreasing IntervalsDerivative Analysis
Function Symmetry
Function symmetry is an important concept when graphing rational functions like \( y = -\frac{1}{x^2} \). Symmetry makes it easier to understand how the graph behaves around the axes. For a function to be symmetric with respect to the y-axis, it must satisfy the condition \( f(-x) = f(x) \). This means that the graph on one side of the y-axis is a mirror image of the graph on the other side.
In this case, substituting \(-x\) into the function results in \( y = -\frac{1}{(-x)^2} = -\frac{1}{x^2} \). We observe that the expression remains unchanged, indicating y-axis symmetry. This type of symmetry shows that the behavior of the function at positive x values is mirrored at the negative x values. This insight simplifies the graphing process, offering a better visual understanding of how the function transforms over its domain.
In this case, substituting \(-x\) into the function results in \( y = -\frac{1}{(-x)^2} = -\frac{1}{x^2} \). We observe that the expression remains unchanged, indicating y-axis symmetry. This type of symmetry shows that the behavior of the function at positive x values is mirrored at the negative x values. This insight simplifies the graphing process, offering a better visual understanding of how the function transforms over its domain.
Increasing and Decreasing Intervals
Identifying intervals where the function is increasing or decreasing provides a deeper insight into its behavior. For the function \( y = -\frac{1}{x^2} \), we find it is undefined at \( x = 0 \), but we can analyze the intervals \((-fty, 0)\) and \((0, fty)\).
The function decreases as the absolute value of \( x \) rises, either positively or negatively. This is because as \( |x| \) increases, the magnitude of \( -\frac{1}{x^2} \) becomes larger and thus more negative. Therefore, we say:
The function decreases as the absolute value of \( x \) rises, either positively or negatively. This is because as \( |x| \) increases, the magnitude of \( -\frac{1}{x^2} \) becomes larger and thus more negative. Therefore, we say:
- The function decreases over the interval \((-fty, 0)\)
- The function also decreases over \((0, fty)\)
Derivative Analysis
Derivatives play a crucial role in determining where a function is increasing or decreasing. By analyzing the derivative, we gain insight into the function's slope at various points in its domain.
For \( y = -\frac{1}{x^2} \), the derivative is calculated as \( y' = \frac{2}{x^3} \). This derivative is positive when \( x < 0 \) and negative when \( x > 0 \). But note that being positive or negative here refers to the position of the graph rather than increase or decrease directions — which is an essential distinction.
As the derivative here shifts signs across \( x = 0 \), the function itself is entirely decreasing, meaning its downward slope becomes steeper as we move away from zero. From a derivative perspective, we conclude:
For \( y = -\frac{1}{x^2} \), the derivative is calculated as \( y' = \frac{2}{x^3} \). This derivative is positive when \( x < 0 \) and negative when \( x > 0 \). But note that being positive or negative here refers to the position of the graph rather than increase or decrease directions — which is an essential distinction.
As the derivative here shifts signs across \( x = 0 \), the function itself is entirely decreasing, meaning its downward slope becomes steeper as we move away from zero. From a derivative perspective, we conclude:
- The negative slope of the function suggests a decreasing behavior
- The change in sign at \( x = 0 \) accentuates the continuous decrease
Other exercises in this chapter
Problem 8
In Exercises 5–30, determine an appropriate viewing window for the given function and use it to display its graph. $$ f(x)=4 x^{3}-x^{4} $$
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Describe the graphs of the equations in Exercises 5–8. $$ x^{2}+y^{2}=0 $$
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