Chapter 5: Problem 75
Find the functions \(f \circ g\) and \(g \circ f\) and their domains. $$ f(x)=2^{x}, \quad g(x)=x+1 $$
Short Answer
Expert verified
\( f \circ g = 2^{x+1} \) and \( g \circ f = 2^{x} + 1 \), both with domains \( (-\infty, \infty) \).
Step by step solution
01
Understand Composition of Functions
To find the composition of two functions, say \( f \) and \( g \), you need to substitute one function into the other. For \( f \circ g \), substitute \( g(x) \) into \( f(x) \). For \( g \circ f \), substitute \( f(x) \) into \( g(x) \).
02
Find \( f \circ g \)
Substitute \( g(x) = x + 1 \) into \( f(x) = 2^{x} \). Therefore, \( f(g(x)) = 2^{g(x)} = 2^{x+1} \). So, \( f \circ g = 2^{x+1} \).
03
Determine Domain of \( f \circ g \)
For \( f \circ g = 2^{x+1} \), the function is defined for all real values of \( x \). Thus, the domain of \( f \circ g \) is all real numbers (\( -\infty, \infty \)).
04
Find \( g \circ f \)
Substitute \( f(x) = 2^{x} \) into \( g(x) = x + 1 \). Therefore, \( g(f(x)) = g(2^{x}) = 2^{x} + 1 \). So, \( g \circ f = 2^{x} + 1 \).
05
Determine Domain of \( g \circ f \)
For \( g \circ f = 2^{x} + 1 \), the function is defined for all real values of \( x \). Thus, the domain of \( g \circ f \) is also all real numbers (\( -\infty, \infty \)).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Domain of a Function
The domain of a function is a crucial concept in mathematics, particularly in function composition. It specifies the set of input values (x-values) for which the function is defined. In simpler terms, it tells us which x values are "allowed" to be plugged into the function, without causing any issues such as division by zero or taking the square root of a negative number.
For the function composition case in the original problem:
For the function composition case in the original problem:
- Domain of \( f \circ g \): Since \( f \circ g = 2^{x+1} \), we note that exponential functions are defined for all real numbers. Therefore, the domain of \( f \circ g \) is all real numbers, expressed mathematically as \( (-\infty, \infty) \).
- Domain of \( g \circ f \): Similarly, \( g \circ f = 2^{x} + 1 \) is a function that defines for all x values on the real number line, meaning the domain is also \( (-\infty, \infty) \).
Exploring Exponential Functions
Exponential functions are powerful mathematical expressions where the variable appears in the exponent. They have the form \( f(x) = a^{x} \), where \( a \) is a positive real number. One of their defining features is their rapid increase or decrease, making them vital in modeling real-world phenomena like population growth or radioactive decay.
In the original exercise, the function \( f(x) = 2^{x} \) is a clear example of an exponential function. Here are some key properties:
In the original exercise, the function \( f(x) = 2^{x} \) is a clear example of an exponential function. Here are some key properties:
- Exponential functions are always positive, because a positive number raised to any power is positive.
- The domain is all real numbers \((-\infty, \infty)\), while the range is positive real numbers \((0, \infty)\).
- An exponential function will always pass through the point \((0,1)\), because any number raised to the power of zero equals one.
Mastering Algebraic Manipulation in Function Composition
Algebraic manipulation is a fundamental skill in solving and understanding function composition. It involves the ability to rearrange, simplify, and substitute expressions in a mathematically valid way. When finding \( f \circ g \) or \( g \circ f \), these skills are crucial.
Let's break it down further using the problem at hand:
Let's break it down further using the problem at hand:
- For \( f \circ g \), we take \( g(x) = x+1 \) and replace x in \( f(x) = 2^x \) with \( x+1 \). The result is \( 2^{x+1} \), effectively shifting the graph of \( 2^x \) one unit to the left.
- For \( g \circ f \), substitute \( f(x) = 2^x \) into \( g(x) = x+1 \), yielding \( 2^x + 1 \). This operation reflects a vertical shift upward by one unit on the graph of \( 2^x \).