Problem 75 - Algebra and Trigonometry

Question

Find the functions \(f \circ g\) and \(g \circ f\) and their domains. $$ f(x)=2^{x}, \quad g(x)=x+1 $$

Step-by-Step Solution

Answer

\( f \circ g = 2^{x+1} \) and \( g \circ f = 2^{x} + 1 \), both with domains \( (-\infty, \infty) \).

Step 1: 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 \( g(x) \).

Step 2: 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, \( f \circ g = 2^{x+1} \).

Step 3: 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 (\( -\infty, \infty \)).

Step 4: 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, \( g \circ f = 2^{x} + 1 \).

Step 5: 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 (\( -\infty, \infty \)).

Key Concepts

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:

• 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 \( (-\infty, \infty) \).

Whether dealing with linear, quadratic, or exponential functions, checking a function's domain ensures that the function is properly defined for all intended inputs.

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:

• 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.

Understanding these properties helps when composing functions like \( f \circ g \) where \( g(x) = x+1 \) is substituted into \( f(x) = 2^x \). After substitution, since the structure of an exponential function does not change, we get the confident result \( f \circ g = 2^{x+1} \).

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:

• 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 \( 2^x \).

By understanding algebraic manipulation, one can efficiently determine composition outcomes and interpret each step categorically. This process aids in visualizing changes in the graph, which is vital in solving real-world problems.