Problem 51 - College Algebra Essentials

Question

In Exercises 51–66, find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=2 x, g(x)=x+7$$

Step-by-Step Solution

Answer

a. \(f \circ g)(x) = 2x + 14\), b. \(g \circ f)(x) = 2x + 7\), c. \(f \circ g)(2) = 18\), d. \(g \circ f)(2) = 11.

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

We first find \(f(g(x))\). Substitute \(g(x)\) into \(f(x)\): \(f(g(x)) = 2(g(x)) = 2(x+7)\)

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

Simplify the equation by distributing 2: \(f(g(x)) = 2x + 14\). Therefore, \(f \circ g)(x) = 2x + 14.\)

Step 3: Finding (g \circ f)(x)

We next find \(g(f(x))\). Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = (f(x)) + 7 = 2x + 7.\)

Step 4: Calculating (f \circ g)(2)

Next is to compute the value of \(f(g(2))\). Substitute x=2 into \(f(g(x)) = 2x + 14\) gives \(f(g(2)) = 2(2) + 14 = 18.\)

Step 5: Calculating (g \circ f)(2)

Finally, compute the value of \(g(f(2))\). Substitute x=2 into \(g(f(x)) = 2x + 7\) gives \(g(f(2)) = 2(2) + 7 = 11.\)

Key Concepts

Composite Functions

Composite functions are the bread and butter of combining two functions into a single operation. Imagine you're on a two-leg journey: first you go to a transit point, and then you continue to your final destination. In math, this journey is like applying one function to the results of another.

In our exercise, we have two functions, \( f(x) = 2x \) and \( g(x) = x + 7 \). To create a composite function, \( f \circ g \), we put \( g(x) \) into \( f(x) \), like giving a relay baton to the second runner. This resulted in \( f(g(x)) = 2(x + 7) \), which simplifies down to \( 2x + 14 \). Similarly, \( g \circ f \) means that f(x) is our first runner, and we put that into \( g(x) \) to get \( g(f(x)) = 2x + 7 \). It's like a mathematical handshake between functions!

Function Operations

Function operations include addition, subtraction, multiplication, and division of functions, much like basic arithmetic with numbers—but instead of numbers, we're working with algebraic expressions. When we combine functions through these operations, we're essentially blending their rules or effects.

In our example, we used function operations to work out the composite functions. When \( f(x) \) operated on the outcome of \( g(x) \) through multiplication (\( f \circ g \) or \( f(g(x)) = 2(g(x)) \) ), we got an entirely new function. Moreover, evaluating these composite functions at specific points, like \( x = 2 \), is a practical application of function operations, helping us see the output of this fusion at particular values.

Algebraic Functions

Algebraic functions serve as puzzles to solve or formulas to put to work. They take our numbers or expressions (inputs), process them through algebraic operations like addition, multiplication, or even powers, and give us a new outcome (output). Most of the functions we encounter in algebra are of this type.

Our textbook problem provides classic examples of algebraic functions: \( f(x) = 2x \) and \( g(x) = x + 7 \). Both are straightforward and involve basic algebraic operations—multiplication and addition, respectively. Understanding these simple yet fundamental building blocks allows us to tackle more complex algebraic functions with confidence.