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Chapter 2: Problem 1

Write the domain of the function in interval notation. $$f(x)=\sqrt{5-2 x}$$

Short Answer

Expert verified
The domain is \( (-\infty, \frac{5}{2}] \).

Step by step solution

01

Title - Identify the constraint for the square root

For the function to be defined, the expression inside the square root must be non-negative. Therefore, set up the inequality: \(5 - 2x \geq 0\)
02

Title - Solve the inequality

Solve the inequality from Step 1 to find the values of \(x\) that satisfy it. First, rearrange the inequality:\(-2x \geq -5\)Next, divide both sides by -2 (and reverse the inequality symbol):\(x \leq \frac{5}{2}\)
03

Title - Write the domain in interval notation

The domain includes all values of \(x\) up to \(\frac{5}{2}\), including \(\frac{5}{2}\) itself. Thus, the domain in interval notation is: \( (-\infty, \frac{5}{2}] \)

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Key Concepts

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

interval notation
Interval notation is a way of expressing the set of all numbers between two endpoints. It uses brackets to show if the endpoints are included or not.
Here鈥檚 a quick breakdown:
  • Square brackets [ ] mean that the endpoint is included in the set.
  • Round brackets ( ) mean that the endpoint is not included in the set.
  • For example, [1, 4) means all numbers between 1 and 4, including 1 but not 4.
In the given exercise, the domain of the function involves all values of x that satisfy the inequality. If x can be any number up to and including \(\frac{5}{2}\), then it鈥檚 written as \( (-\text{鈭瀩, \frac{5}{2}] \). This notation means x can be any number from negative infinity to \(\frac{5}{2}\), including \( \frac{5}{2} \).
square root function
A square root function involves the square root of a variable, written as \( \sqrt{...} \). It's important to recognize that for a square root function to be real-valued (i.e., only producing real numbers as outputs), the expression inside the square root must be non-negative (0 or positive).
  • For instance, in the function given, f(x) = \( \sqrt{5-2x} \), the expression inside the square root is \( 5 - 2x \). For this square root to be defined, \( 5 - 2x \) must be non-negative.
  • This condition translates into the inequality \( 5 - 2x \geq 0 \).
When dealing with square root functions, always start by identifying this kind of inequality to determine the domain of the function.
inequality
Inequalities compare two values or expressions and tell us if one is greater than, less than, or equal to the other. In this problem, the inequality \(5 - 2x \geq 0 \) is used to determine the domain of the function.
Here鈥檚 how to solve it step-by-step:
  • Start with the inequality \( 5 - 2x \geq 0 \).
  • Subtract 5 from both sides to isolate the term with x: \( -2x \geq -5 \).
  • Divide both sides by -2. Remember, dividing or multiplying an inequality by a negative number reverses the inequality symbol!
So, after dividing by -2, you get: \( x \leq \frac{5}{2} \).
This tells you the function is defined for all x-values less than or equal to \(\frac{5}{2}\). Writing this in interval notation gives the interval: \( (-\text{鈭瀩, \frac{5}{2}] \).

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Most popular questions from this chapter

Refer to the functions \(f, g,\) and \(h\) and evaluate the given functions. \(f(x)=2 x+1 \quad g(x)=x^{2} \quad h(x)=\sqrt[3]{x}\) $$(g \circ h \circ f)(x)$$

Use a graphing utility to graph the piecewise-defined function. $$ \begin{aligned} &z(x)=\left\\{\begin{array}{ll} 2.5 x+8 & \text { for } x<-2 \\ -2 x^{2}+x+4 & \text { for }-2 \leq x<2 \\ -2 & \text { for } x \geq 2 \end{array}\right.\\\ &\text { Is there actually a "gap" in the graph at } x=2 ? \end{aligned} $$

Determine if the function is even, odd, or neither. $$ h(x)=5 x $$

Use a graphing utility to graph the piecewise-defined function. $$ \begin{aligned} &k(x)=\left\\{\begin{array}{ll} -2.7 x-4.1 & \text { for } x \leq-1 \\ -x^{3}+2 x+5 & \text { for }-1

Graph the function. $$ s(x)=\left\\{\begin{array}{ll} -x-1 & \text { for } x \leq-1 \\ \sqrt{x+1} & \text { for } x>-1 \end{array}\right. $$
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