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Magazine Chapter 8: Problem 79
Find the domain of each function. $$ f(x)=\sqrt{x^{2}+8 x} $$
Short Answer
Expert verified
Domain: \( (-\infty, -8] \cup [0, \infty) \).
Step by step solution
01
Identify the function and restriction
Given the function is \( f(x) = \sqrt{x^{2}+8x} \). The domain is the set of all x-values for which the function is defined. Since this is a square root function, the expression inside the square root must be non-negative.
02
Set up the inequality
Set up the inequality for the expression inside the square root to be greater than or equal to zero: \( x^2 + 8x \geq 0 \).
03
Solve the inequality
To solve the inequality \( x^2 + 8x \geq 0 \), find the roots of the corresponding equation \( x^2 + 8x = 0 \). Factorize the quadratic equation to get: \( x(x + 8) = 0 \). Therefore, the roots are \( x = 0 \) and \( x = -8 \).
04
Determine the intervals
Use the roots to divide the real number line into intervals: \( (-\infty, -8) \), \( (-8, 0) \), and \( (0, \infty) \). Test a point in each interval to determine where the inequality holds: For \( x = -9 \) in \( (-\infty, -8) \), \((-9)^2 + 8(-9) = 81 - 72 = 9 < 0 \). For \( x = -4 \) in \( (-8, 0) \), \((-4)^2 + 8(-4) = 16 - 32 = -16 < 0 \). For \( x = 1 \) in \( (0, \infty) \), \((1)^2 + 8(1) = 1 + 8 = 9 > 0 \).
05
Combine results
The inequality \( x^2 + 8x \geq 0 \) holds in the intervals where the expression is non-negative. The valid intervals are \( (-\infty, -8] \cup [0, \infty) \), which is the domain of the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Inequality
A quadratic inequality involves an expression of the form \( ax^2 + bx + c \) being greater than or less than a certain value.
In our exercise, we are dealing with the inequality \( x^2 + 8x \geq 0 \).
The general steps to solve quadratic inequalities are:
In our exercise, we are dealing with the inequality \( x^2 + 8x \geq 0 \).
The general steps to solve quadratic inequalities are:
- First, find the roots of the equation by solving \( ax^2 + bx + c = 0 \). For our quadratic equation, the roots are found by factorizing \( x(x + 8) = 0 \), giving roots \( x = 0 \) and \( x = -8 \).
- Next, divide the number line into intervals based on these roots: \( (-\infty, -8) \), \( (-8, 0) \), \( (0, \infty) \).
- Then, choose test points in each interval to determine where the inequality holds true. For example, test \( x = -9 \) in \( (-\infty, -8) \) and you get \( 81 - 72 = 9 > 0 \), meaning this interval does not satisfy the inequality. Repeat for \( x = -4 \) and \( x = 1 \).
- Finally, combine the positive intervals. For \( x^2 + 8x \geq 0 \), the valid intervals are \( (-\infty, -8] \cup [0, \infty) \).
Square Root Function
The square root function is written as \( f(x) = \sqrt{g(x)} \). For this function to be defined, the expression inside the square root, \( g(x) \), must be non-negative (i.e., \( g(x) \geq 0 \)).
In our exercise, given \( f(x) = \sqrt{x^2 + 8x} \), we need to ensure that \( x^2 + 8x \geq 0 \).
The function is defined only for the non-negative values of \( x^2 + 8x \), as taking the square root of a negative number results in a complex value, which is outside the scope of real-valued functions.
In our exercise, given \( f(x) = \sqrt{x^2 + 8x} \), we need to ensure that \( x^2 + 8x \geq 0 \).
The function is defined only for the non-negative values of \( x^2 + 8x \), as taking the square root of a negative number results in a complex value, which is outside the scope of real-valued functions.
- Setting up the inequality \( x^2 + 8x \geq 0 \) is critical because it ensures the expression under the square root yields non-negative values.
- Solving the inequality by finding the roots (critical points) and testing the intervals identifies where the square root function is valid.
Interval Notation
Interval notation is a way to express the set of all numbers between two endpoints. For example, \( (a, b) \) represents all numbers between \( a \) and \( b \) but not including \( a \) and \( b \), while \( [a, b] \) includes both endpoints.
In our exercise, solving the inequality \( x^2 + 8x \geq 0 \) provided the intervals \( -\infty, -8 \) and \( 0, \infty \).
Using interval notation:
In our exercise, solving the inequality \( x^2 + 8x \geq 0 \) provided the intervals \( -\infty, -8 \) and \( 0, \infty \).
Using interval notation:
- \( (-\infty, -8) \) is transformed to \( (-\infty, -8] \) because \( x = -8 \) is included in the valid intervals.
- \( (0, \infty) \) is transformed to \( [0, \infty) \) because \( x = 0 \) is also included.
- The combined interval is \( (-\infty, -8] \cup [0, \infty) \).
