Problem 8 - 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

f(g(3)) = -14, g(f(3)) = -19. The composition of functions is not commutative.

Step 1: Find g(3)

First, substitute 3 into the function \(g(x)\). The expression becomes \(g(3) = 1 - 2(3)\). Calculate the expression: \(1 - 6 = -5\). So, \(g(3) = -5\).

Step 2: Find f(g(3))

Now, take the result from Step 1 and substitute it into the function \(f(x)\). The expression becomes \(f(-5) = 3(-5) + 1\). Calculate the expression: \(-15 + 1 = -14\). Thus, \(f(g(3)) = -14\).

Step 3: Find f(3)

Next, substitute 3 into the function \(f(x)\). The expression becomes \(f(3) = 3(3) + 1\). Calculate the expression: \(9 + 1 = 10\). So, \(f(3) = 10\).

Step 4: Find g(f(3))

Now, take the result from Step 3 and substitute it into the function \(g(x)\). The expression becomes \(g(10) = 1 - 2(10)\). Calculate the expression: \(1 - 20 = -19\). Thus, \(g(f(3)) = -19\).

Step 5: Compare f(g(3)) and g(f(3))

Observe the results from Step 2 and Step 4: \(f(g(3)) = -14\) and \(g(f(3)) = -19\). Since \(f(g(3))\) is not equal to \(g(f(3))\), the composition of functions is not commutative.

Key Concepts

Algebraic Functions

Algebraic functions are expressions that involve polynomial equations, often taken in terms of variables like \(x\). For instance, in the given exercise, we have two functions: \(f(x) = 3x + 1\) and \(g(x) = 1 - 2x\). These are examples of linear functions, which is a key type within algebraic functions.
These functions serve as versatile tools in mathematics, enabling you to model relationships between variables. When working with algebraic functions, remember the basics:

• Identify the coefficients and constants, such as 3 and 1 in \(f(x) = 3x + 1\).
• Recognize the operations involved, here multiplication and addition in \(f\), and subtraction and multiplication in \(g\).
• To evaluate these functions, substitute values as seen in the steps where 3 was substituted into both \(f\) and \(g\).

By thoroughly understanding these basics, you can evaluate and analyze more complex algebraic expressions confidently.

Commutative Property

The commutative property is fundamental in algebra, stating that position does not change the result for operations like addition and multiplication. For example, \(a + b = b + a\) or \(ab = ba\). However, when dealing with function composition, we're looking at a different scenario.
Function composition, represented by \((f \circ g)(x)\), means first applying \(g(x)\), then \(f(x)\) to the result, and vice versa for \((g \circ f)(x)\). In the original exercise, it was observed that \(f(g(3)) = -14\) and \(g(f(3)) = -19\).
These results show that function composition is not commutative, as flipping the order of functions can lead to different results. Always keep this difference in mind, especially when applying operations to solve equations or when analyzing sequential processes.

Intermediate Algebra

Intermediate algebra goes beyond basic concepts and introduces more complex functionalities such as operations on functions—like composition. Being able to compose functions is an important skill. It requires understanding both the individual functions and how they interact when applied in succession.
In the exercise, we saw this with \(f(x)\) and \(g(x)\). Intermediate algebra also involves comprehending the significance of sequential function application. For example, solving \(f(g(3))\) involves several steps:

• Calculate \(g(3)\), resulting in \(-5\).
• Use \(-5\) as input for \(f(x)\) to find \(f(-5)\), leading to \(-14\).

Despite these being seemingly basic steps, these layered tasks form the backbone of intermediate algebra, demanding attention to detail and logical thinking.
By mastering these skills, students can tackle more advanced problems, enhance their mathematical reasoning, and prepare for further algebra study.