Problem 48
Question
Find the domain of the function. $$ g(x)=\sqrt{7-3 x} $$
Step-by-Step Solution
Verified Answer
The domain of the function is \((-\infty, \frac{7}{3}]\).
1Step 1: Understanding the Function
The given function is a square root function, which is defined as \( g(x) = \sqrt{7 - 3x} \). A square root function is defined only when the expression under the square root is non-negative.
2Step 2: Setting Up the Inequality
To find the domain of \( g(x) \), determine when the expression under the square root is greater than or equal to zero: \( 7 - 3x \geq 0 \).
3Step 3: Solving the Inequality
Solve the inequality \( 7 - 3x \geq 0 \). Start by subtracting 7 from both sides to get \( -3x \geq -7 \). Then, divide both sides by -3, reversing the inequality sign, to get \( x \leq \frac{7}{3} \).
4Step 4: Expressing the Domain
The solution to the inequality \( x \leq \frac{7}{3} \) describes the domain of the function. Therefore, the domain of \( g(x) \) is all real numbers \( x \) such that \( x \leq \frac{7}{3} \). In interval notation, this is expressed as \( (-\infty, \frac{7}{3}] \).
Key Concepts
Square Root FunctionInequality SolvingInterval Notation
Square Root Function
Square root functions are a specific type of function involving the square root of an expression. The general form of a square root function is \( f(x) = \sqrt{a} \), where \( a \) is called the radicand. The primary rule governing these functions is that the radicand must be non-negative to ensure the function is defined in the real number system.
- A positive radicand allows a real number output, while a negative radicand would yield an imaginary number result, not considered in its domain within real numbers.
- This constraint informs us to focus on solving inequality problems where a \( \geq 0 \) (greater than or equal to zero) is aimed.
Inequality Solving
Solving inequalities is crucial when finding the domain of functions involving square roots. Inequalities help identify the range of values where the function remains real. Here's a simple guide on how we solve the inequality step by step.
- Subtract 7 from both sides to get \( -3x \ge -7 \). Remember!- When dividing by a negative number, the inequality sign reverses, making \( x \le \frac{7}{3} \).
This solution tells us the allowable values for \( x \) that make the entire expression under the square root non-negative.
- Begin with setting up an inequality based on the requirement that the radicand must be non-negative, for example, \( 7 - 3x \ge 0 \).
- To solve, isolate \( x \) by performing arithmetic operations until \( x \) is on one side of the inequality.
- Subtract 7 from both sides to get \( -3x \ge -7 \). Remember!- When dividing by a negative number, the inequality sign reverses, making \( x \le \frac{7}{3} \).
This solution tells us the allowable values for \( x \) that make the entire expression under the square root non-negative.
Interval Notation
Interval notation is a concise way to express the domain or range of a function. It represents all the possible values a function can take under certain conditions and involves brackets and parentheses. Understanding how to interpret interval notation is key:
- Parentheses \(( )\) denote that an endpoint is not included in the interval.
- Brackets \([ ]\) indicate the endpoint is included.
- An interval like \((-\infty, \frac{7}{3}]\) defines all numbers less than or equal to \( \frac{7}{3} \).
Other exercises in this chapter
Problem 48
Sketch the graph of the piecewise defined function. $$ f(x)=\left\\{\begin{array}{ll}{x^{2}} & {\text { if }|x| \leq 1} \\ {1} & {\text { if }|x|>1}\end{array}\
View solution Problem 48
33–48 ? Sketch the graph of the function, not by plotting points, but by starting with the graph of a standard function and applying transformations. $$ y=2-|x|
View solution Problem 49
\(45-50\) Express the function in the form \(f \circ g\) $$ H(x)=\left|1-x^{3}\right| $$
View solution Problem 49
Sketch the graph of the piecewise defined function. $$ f(x)=\left\\{\begin{array}{ll}{4} & {\text { if } x2}\end{array}\right. $$
View solution