Problem 19 - Beginning and Intermediate Algebra

Question

Let \(f(x)=5 x-4\) and \(g(x)=x+7 .\) Find a) \((f \circ g)(x)\) b) \(\quad(g \circ f)(x)\) c) \((f \circ g)(3)\)

Step-by-Step Solution

Answer

a) \((f \circ g)(x) = 5x + 31\) b) \((g \circ f)(x) = 5x + 3\) c) \((f \circ g)(3) = 46\)

Step 1: Find \((f\circ g)(x)\)

To find the composition \((f\circ g)(x)\), we need to substitute the function \(g(x)\) into \(f(x)\). In other words, we have to replace the variable \(x\) in the expression of \(f(x)\) with the expression of \(g(x)\). So, we have: \((f\circ g)(x) = f(g(x)) = 5(g(x)) - 4\) Now, let's substitute the expression of \(g(x)\) into the equation: \((f\circ g)(x) = 5(x + 7) - 4\) After simplifying, we get: \((f\circ g)(x) = 5x + 35 - 4\) \((f\circ g)(x) = 5x + 31\)

Step 2: Find \((g\circ f)(x)\)

Now, let's find the composition \((g\circ f)(x)\). Similarly to the previous step, we need to substitute the function \(f(x)\) into \(g(x)\). So, we have: \((g\circ f)(x) = g(f(x)) = (5x - 4) + 7\) Now, let's simplify the expression: \((g\circ f)(x) = 5x - 4 + 7\) \((g\circ f)(x) = 5x + 3\)

Step 3: Calculate \((f\circ g)(3)\)

Finally, we need to find the value of the composition \((f\circ g)\) at the point \(x = 3\). We have already found the expression for \((f\circ g)(x)\) in step 1, so we can substitute the value of \(x = 3\) into the equation: \((f\circ g)(3) = 5(3) + 31\) \((f\circ g)(3) = 15 + 31\) \((f\circ g)(3) = 46\) So, the answers to the exercise are: a) \((f \circ g)(x) = 5x + 31\) b) \((g \circ f)(x) = 5x + 3\) c) \((f \circ g)(3) = 46\)

Key Concepts

Function Composition

In algebra, function composition is the process of applying one function to the results of another. It can be thought of as placing one function inside of another like a giant puzzle. It allows us to combine functions and use the output of one as the input of the other.
In notation, the composition of two functions, such as \(f(x)\) and \(g(x)\), is written as \((f \circ g)(x)\). This means you first apply \(g(x)\) to \(x\), and then apply \(f(x)\) to the result.
For example, if we have \(f(x) = 5x - 4\) and \(g(x) = x + 7\), we find \((f \circ g)(x)\) by replacing every \(x\) in \(f(x)\) with \(g(x)\). Thus, we compute it as \(f(g(x)) = 5(g(x)) - 4\), which ultimately gives us the expression \(5x + 31\) after simplification.
Essentially, function composition transforms one function's output into the starting point for another function, allowing for a seamless calculation of complex processes.

Polynomial Functions

Polynomial functions are a type of algebraic expression consisting of variables raised to whole number powers and multiplied by coefficients. They can range from simple linear equations to intricate multi-variable systems.
Polynomial functions follow the form \(a_n x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\), where \(a_n, a_{n-1}, ..., a_1, a_0\) are constants.

• The degree of a polynomial is determined by the highest power of \(x\) in the equation.
• In our example, both \(f(x) = 5x - 4\) and \(g(x) = x + 7\) are first-degree polynomials, making them linear expressions.
• These functions form the backbone of many algebraic processes, including function composition, as seen through the combination of \((f \circ g)(x)\) to yield a new polynomial from existing ones.

Understanding polynomial functions provides a strong foundation for delving into broader algebraic topics and solving real-world problems.

Input-Output Analysis

Input-output analysis in algebra focuses on understanding how changing the input of a function affects its output. This process is crucial when dealing with function compositions or any other algebraic manipulations.
By examining an expression like \((f \circ g)(x)\), we see how the input \(x\) initially operates through \(g(x)\) to become \(x + 7\), which then serves as the input for \(f(x)\). The overall effect calculates \((f \circ g)(x) = 5(x + 7) - 4\) by treating \(g(x)\) as an input for \(f(x)\).

• This results in a straightforward output of \(5x + 31\), as \(x+7\) shapes the outcome when put into \(f(x)\).
• Conducting the analysis for a specific value, like \((f \circ g)(3)\), highlights how the input is transformed step by step into a precise output, in this case, 46.

By tracking the journey from input through each function layer, algebraic solutions become logical and less cumbersome.