Problem 8 - Elementary and Intermediate Algebra

Question

a. If \(f(x)=3 x+1\) and \(g(x)=1-2 x,\) find \(f(g(3))\) and \(g(f(3))\) b. Is the composition of functions commutative?

Step-by-Step Solution

Answer

a) \(f(g(3)) = -14\), \(g(f(3)) = -19\).\n b) No, the composition of functions is not commutative.

Step 1: Find g(3)

We need to evaluate the function \(g(x) = 1 - 2x\) at \(x = 3\). Substitute 3 into the expression: \[g(3) = 1 - 2(3) = 1 - 6 = -5.\]

Step 2: Find f(g(3))

Now substitute the value of \(g(3)\), which is \(-5\), into the function \(f(x) = 3x + 1\). Calculate \[f(g(3)) = f(-5) = 3(-5) + 1 = -15 + 1 = -14.\]

Step 3: Find f(3)

Evaluate the function \(f(x) = 3x + 1\) at \(x = 3\). Substitute 3 into the expression: \[f(3) = 3(3) + 1 = 9 + 1 = 10.\]

Step 4: Find g(f(3))

Substitute the value of \(f(3)\), which is \(10\), into the function \(g(x) = 1 - 2x\). Calculate \[g(f(3)) = g(10) = 1 - 2(10) = 1 - 20 = -19.\]

Step 5: Analyze Commutativity

Check if \(f(g(3)) = g(f(3))\). We found \(f(g(3)) = -14\) and \(g(f(3)) = -19\). Since \(-14 eq -19\), the composition of functions \(f(g(x))\) and \(g(f(x))\) are not equal and hence, not commutative.

Key Concepts

Commutativity of Functions

In mathematics, the term "commutative" often refers to the ability to swap two elements' positions without affecting the outcome. For example, in arithmetic, addition is commutative because \(a + b = b + a\). However, when we consider functions, this is not always the case. Function composition refers to the process of applying one function to the result of another. This process is generally not commutative.

• For two functions \(f\) and \(g\), the composition \(f(g(x))\) involves applying \(g\) first and then \(f\).
• The opposite, \(g(f(x))\), involves applying \(f\) first and then \(g\).

In the exercise example, it's clear that \(f(g(x)) eq g(f(x))\). This non-equality shows that the order of applying functions matters, which highlights the non-commutative nature of function composition.

Evaluation of Functions

Evaluating functions is the process of finding the output for a specific input. In simpler terms, it's like plugging a number into an equation to see what result you get.
Let's look at how this works for functions given in the exercise:

• For \(g(x) = 1 - 2x\), we evaluate it at \(x = 3\), substituting 3 wherever \(x\) appears in the expression, yielding \(g(3) = 1 - 2(3) = -5\).
• For \(f(x) = 3x + 1\), evaluate at \(x = 3\) giving \(f(3) = 3(3) + 1 = 10\).

By carrying out these evaluations, we can assess what the function produces for the given input.
This process is crucial in understanding and working with real-world problems modeled by mathematical functions.

Algebraic Functions

Algebraic functions are expressions constructed using arithmetic operations and variables raised to powers. They form the backbone of much of algebra and are used to model everything from simple equations to complex curves.
In this example, both given functions are algebraic:

• The function \(f(x) = 3x + 1\) is a simple linear function. It involves multiplication and addition, typical operations for an algebraic function.
• The function \(g(x) = 1 - 2x\) similarly combines subtraction and multiplication, thus also being algebraic.

Understanding these functions helps us manipulate their algebraic expressions and solve equations that describe a wide array of phenomena, from physical systems to economic models.