Chapter 2: Problem 25
For each pair of functions, (a) find \((f+g)(x),(f-g)(x),\) and \((f g)(x)\) (b) give the domains of the functions in part (a); (c) find \(\frac{f}{8}\) and give its domain; (d) find \(f \circ g\) and give its domain; and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=|x+3|, g(x)=2 x$$
Short Answer
Step by step solution
Find (f+g)(x)
Find (f-g)(x)
Find (f g)(x)
Domain of (f+g)(x), (f-g)(x), (f g)(x)
Find \(\frac{f}{8}\) and give its domain
Find \(f \circ g\) and give its domain
Find \(g \circ f\) and give its domain
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of Functions
- \( f(x) = |x+3| \)
- \( g(x) = 2x \)
When performing operations like addition, subtraction, or multiplication on these functions, such as \((f+g)(x), (f-g)(x), \text{or} \ (fg)(x)\), the domain remains unchanged. This is because there are no additional operations that would restrict the possible inputs further. Thus, the domain for these combined operations is also \((-\infty, \infty)\). When dividing, such as finding \(\frac{f}{8}(x)\), we again face no further constraints since 8 is a non-zero constant, preserving the domain of all real numbers.
Composition of Functions
- \(f \circ g (x) = f(g(x)) = |2x + 3|\)
- \(g \circ f (x) = g(f(x)) = 2|x + 3|\)
In the reverse composition \(g \circ f\), we apply \(f\) first, resulting in \(|x+3|\), and then this result is multiplied by 2 in \(g\), giving us \(2|x + 3|\). Again, no new domain restrictions are introduced, so the domain of \(g \circ f\) also remains \((-\infty, \infty)\). Function composition thus broadens possibilities while retaining domain integrity when dealing with functions defined on all real numbers.
Absolute Value Function
- For inputs where \(x + 3 \geq 0\), the output is simply \(x + 3\).
- For inputs where \(x + 3 < 0\), the output is \( -(x + 3) \).
Because the absolute value function is designed to handle all real-number inputs without exceptions, its domain naturally extends across the entire real number line, \((-\infty, \infty)\), as indicated in the previous discussions of domain. Understanding \(|x|\) and its resulting effects in algebra and calculus can yield insights into function behaviors and is invaluable for grasping more complex functions and operations.