/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 You are given a pair of function... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given a pair of functions, \(f\) and \(g\). In each case, estimate the domain of \((f+g)(x)\) using your computer or graphing calculator. Confirm analytically. $$ f(x)=\sqrt{2 x+9}, g(x)=\sqrt{2 x-5} $$

Short Answer

Expert verified
The domain of \((f+g)(x)\) is \(x \geq \frac{5}{2}\).

Step by step solution

01

Understand the Domain of Square Root Functions

The domain of a square root function \( f(x) = \sqrt{x} \) is all \( x \geq 0 \) because the square root of a negative number is undefined in the set of real numbers.
02

Find the Domain of \( f(x) = \sqrt{2x+9} \)

To find the domain of \( f(x) = \sqrt{2x+9} \), set the expression inside the square root to be greater than or equal to 0: \( 2x + 9 \geq 0 \). Solving for \( x \), we get \( 2x \geq -9 \) which simplifies to \( x \geq -\frac{9}{2} \). Thus, the domain of \( f(x) \) is \( x \geq -\frac{9}{2} \).
03

Find the Domain of \( g(x) = \sqrt{2x-5} \)

To find the domain of \( g(x) = \sqrt{2x-5} \), set the expression inside the square root to be greater than or equal to 0: \( 2x - 5 \geq 0 \). Solving for \( x \), we get \( 2x \geq 5 \) which simplifies to \( x \geq \frac{5}{2} \). Thus, the domain of \( g(x) \) is \( x \geq \frac{5}{2} \).
04

Determine the Domain of \( (f+g)(x) \)

Since \( (f+g)(x) = f(x) + g(x) \), the domain of \( (f+g)(x) \) is the intersection of the domains of \( f(x) \) and \( g(x) \). This means \( x \) must satisfy both \( x \geq -\frac{9}{2} \) and \( x \geq \frac{5}{2} \). The range satisfying both conditions is \( x \geq \frac{5}{2} \).
05

Confirm Analytically

Check if there are any additional constraints that could further limit the domain by verifying endpoints graphically or numerically, and confirm there are no such constraints beyond \( x \geq \frac{5}{2} \).

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Key Concepts

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

Square Root Functions
When it comes to square root functions, the most important thing to know is their domain. A square root function, like \( f(x) = \sqrt{x} \), is defined only for non-negative numbers.
This is because the square root of a negative number is not defined within the set of real numbers, which you'll commonly work with. To ensure a function is defined, we only consider inputs (values of \( x \)) that make what’s inside the square root positive or zero. For example, if we look at \( f(x) = \sqrt{2x+9} \), we set \( 2x + 9 \geq 0 \) and solve this inequality to identify the domain of \( f(x) \). The understanding of square root functions comes handy when designing or analyzing any algorithm or function that involves these roots. Always ensure what’s inside stays positive as part of defining the validity of any input.
Domain Analysis
Domain analysis is key for determining where a function is "legal." The domain of a function refers to all the possible values of \( x \) that will not cause the function to be undefined.
When analyzing functions, especially those involving square roots, count on inequalities to help you. For \( f(x) = \sqrt{2x+9} \), the possible \( x \) values must satisfy \( 2x + 9 \geq 0 \), which simplifies to \( x \geq -\frac{9}{2} \). As for \( g(x) = \sqrt{2x-5} \), the inequality \( 2x - 5 \geq 0 \) gives us a domain of \( x \geq \frac{5}{2} \). By understanding these definitions, you can determine where a function operates smoothly without running into any undefined expressions. Keeping track of these domains helps manage complex functions with multiple components.
Function Intersection
Function intersection, when it comes to domains, refers to finding the overlap of domains from two or more functions.
This is especially useful when combining functions, such as adding, as in \((f+g)(x) = f(x) + g(x)\). To correctly establish the domain of \((f+g)(x)\), you need to identify where both \( f(x) \) and \( g(x) \) are valid. In other words, \( x \) must meet the criteria of both domains. In our example, \( f(x) \) is defined for \( x \geq -\frac{9}{2} \), while \( g(x) \) requires \( x \geq \frac{5}{2} \). The intersection of these two domains, i.e., the values of \( x \) that satisfy both conditions, is \( x \geq \frac{5}{2} \). The concept of intersection is frequently employed across various mathematical applications, ensuring that combinations of functions operate under all necessary conditions.

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Most popular questions from this chapter

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