Chapter 5: Problem 3
For each function. \(\bullet\) Find its domain. \(\bullet\) Create a sign diagram. \(\bullet\) Use your calculator to help you sketch its graph and identify any vertical or horizontal asymptotes, 'unusual steepness' or cusps. $$f(x)=x \sqrt{1-x^{2}}$$
Short Answer
Expert verified
Domain is \([-1, 1]\); function is negative from \([-1, 0)\), zero at \(-1, 0, 1\), positive at \((0, 1)\). No asymptotes.
Step by step solution
01
Find the Domain
The function is given by \( f(x) = x \sqrt{1-x^2} \). For this function to be defined, the expression under the square root, \( 1-x^2 \), must be equal to or greater than zero. This leads to the inequality \( 1-x^2 \geq 0 \). Solving for \( x \), we get:\[1-x^2 \geq 0 \implies 1 \geq x^2 \implies -1 \leq x \leq 1\]Thus, the domain of \( f(x) \) is \( [-1, 1] \).
02
Create a Sign Diagram
To create a sign diagram for \( f(x) = x \sqrt{1-x^2} \), identify where the function is positive, negative, or zero. The function is zero at the endpoints of the domain and also when \( x = 0 \). Therefore:- When \( x = -1 \), \( f(-1) = 0 \).- For \( -1 < x < 0 \), \( f(x) \) is negative because both \( x \) and \( \sqrt{1-x^2} \) are negative.- At \( x = 0 \), \( f(0) = 0 \).- For \( 0 < x < 1 \), \( f(x) \) is positive because both \( x \) and \( \sqrt{1-x^2} \) are positive.- When \( x = 1 \), \( f(1) = 0 \).This gives a sign diagram: Negative on \([-1, 0)\), zero at \(-1, 0, \text{and } 1\), and positive on \((0, 1)\).
03
Analyze Graph and Asymptotes
To sketch the graph of \( f(x) = x \sqrt{1-x^2} \), note the following:- As identified in Step 1, the function is only defined from \(-1 \leq x \leq 1\).- At \( x = -1, 0, \text{ and } 1 \), the function value is 0, so these are points on the \( x \)-axis.- The graph will curve from the \(-1, 0\) section, touching the \( x \)-axis at \( x = 0 \) and then above for \( 0 < x < 1 \).- There are no vertical or horizontal asymptotes since the function relies on boundaries and is not defined outside of \(-1 \leq x \leq 1\).
04
Determine Unusual Features
Based on the sign diagram and analysis:
- There is no unusual steepness or cusps. The function transitions smoothly as it is derived from the square root of a quadratic expression, creating a semi-ellipse shape in the defined domain.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domains of Functions
In the context of functions, the **domain** refers to the set of all possible input values (or values of the independent variable) that will produce a valid output. For a function like \( f(x) = x \sqrt{1-x^2} \), the key consideration is the expression under the square root. The function is only defined when the square root has a non-negative value.
Therefore, we solve the inequality \(1-x^2 \geq 0 \). After rearranging, we see that \( x^2 \leq 1 \), which implies that \( x \) must lie within the interval \([-1, 1]\) for the function to be defined. This means our domain is \([-1, 1]\). Thus, understanding the domain is crucial because it tells us for which values of \(x\) the function exists.
Therefore, we solve the inequality \(1-x^2 \geq 0 \). After rearranging, we see that \( x^2 \leq 1 \), which implies that \( x \) must lie within the interval \([-1, 1]\) for the function to be defined. This means our domain is \([-1, 1]\). Thus, understanding the domain is crucial because it tells us for which values of \(x\) the function exists.
Asymptotes
Asymptotes are lines that a graph approaches but never actually reaches. They often appear in functions involving fractions or logarithms. There are two common types:
- **Vertical Asymptotes**: These occur when the function approaches infinity or negative infinity as \( x \) nears a specific value.
- **Horizontal Asymptotes**: These appear when \( y \) approaches a constant value as \( x \) goes towards infinity or negative infinity.
Sign Diagrams
A **sign diagram** helps us visualize where a function is positive, negative, or zero. It's like a roadmap of the function's behavior within its domain.
In our function, \( f(x) = x \sqrt{1-x^2} \), we find:
In our function, \( f(x) = x \sqrt{1-x^2} \), we find:
- The function is zero at \( x = -1, 0, \text{ and } 1 \).
- For \( -1 < x < 0 \), the function is negative because \( x \) is negative and the square root is positive, making the product negative.
- For \( 0 < x < 1 \), \( x \) and the square root are both positive, so the product is positive.
Function Graphing
When creating a graph of a function, we integrate understanding from the domain, sign diagram, and potential asymptotes. With \( f(x) = x \sqrt{1-x^2} \), here’s how the graph takes shape:
- The domain \([-1, 1]\) confines the graph to this interval on the x-axis.
- There are important intercepts at \( x = -1, 0, \text{ and } 1 \), all touching the x-axis.
- The lack of asymptotes ensures the function smoothly transitions without infinitely approaching any specific line.
- The shape of the graph mimics a semi-ellipse due to the structure of the square root, concave at the origin.