Problem 69 - College Algebra with Modeling and Visualization

Question

Exercises \(57-72:\) Use the given \(f(x)\) and \(g(x)\) to find each of the following. Identify its domain. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=1-5 x, \quad g(x)=\frac{1-x}{5} $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = x\), (b) \((g \circ f)(x) = x\), (c) \((f \circ f)(x) = 25x - 4\); Domains: \((- fty, fty)\) for all.

Step 1: Understanding Function Composition

Function composition involves plugging one function into another. For (a) \((f \circ g)(x)\) means plugging \(g(x)\) into \(f(x)\), for (b) \((g \circ f)(x)\) means plugging \(f(x)\) into \(g(x)\), and for (c) \((f \circ f)(x)\) means plugging \(f(x)\) into itself.

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

Substitute \(g(x)\) into \(f(x)\): \(f(g(x)) = f\left(\frac{1-x}{5}\right) = 1 - 5\left(\frac{1-x}{5}\right)\). Simplify this expression: \(1 - (1 - x) = x\). Thus, \((f \circ g)(x) = x\).

Step 2: Domain of \((f \circ g)(x)\)

Since \((f \circ g)(x) = x\), it is defined for all real numbers, just like the variable \(x\). Thus, the domain is \((-fty, fty)\).

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

Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = g(1-5x) = \frac{1-(1-5x)}{5} = \frac{5x}{5}\). Simplify this to get \((g \circ f)(x) = x\).

Step 4: Domain of \((g \circ f)(x)\)

Since \((g \circ f)(x) = x\), it is defined for all real numbers. Thus, the domain is \((-fty, fty)\).

Step 5: Calculate \((f \circ f)(x)\)

Substitute \(f(x)\) into itself: \(f(f(x)) = f(1-5x) = 1 - 5(1-5x)\). Simplify: \(1 - 5 + 25x = 25x - 4\). Thus, \((f \circ f)(x) = 25x - 4\).

Step 6: Domain of \((f \circ f)(x)\)

Since \((f \circ f)(x) = 25x - 4\) is a polynomial, it is defined for all real numbers. Thus, the domain is \((-fty, fty)\).

Key Concepts

Domain of a Function

When we talk about the "domain of a function," we're referring to all the possible input values (usually represented by \(x\)) that allow the function to work without any issues. For most regular algebraic functions, like polynomials, this is pretty straightforward. They work for any real number you throw at them.

However, some functions have restrictions. For example, you can't take the square root of a negative number if you're only dealing with real numbers, or divide by zero—these actions aren't defined in the realm of real numbers when working with standard arithmetic. That's why it's always important to check for potential issues.

• Take a function like \(f(x) = 25x - 4\). This is a polynomial, and it includes all real numbers in its domain because there's no division by zero or square roots to worry about.
• If we had a function like \(g(x) = \frac{1}{x}\), the domain wouldn't include \(x = 0\) since we'd be dividing by zero.

Real Numbers

Real numbers are essentially all the numbers that exist on the number line. This includes:

• Whole numbers like 0, 1, 2, etc.
• Fractions and decimals, such as 3.14 or \(\frac{1}{2}\).
• Negative numbers like -5 or -\(\frac{3}{4}\).

Real numbers are crucial when you're talking about functions because they represent all potential "realistic" values and results you can deal with in everyday calculations. The domain of typical functions is usually a subset of real numbers.

In exercises dealing with function composition, like our \(f \circ g\)(x) or \(g \circ f\)(x), an understanding of real numbers helps us quickly see why these compositions are defined across all real numbers. Put simply, they don't involve any operations like division by zero that would make them undefined for certain numbers.

Function Simplification

Function simplification is about making complex functions easier to handle. It involves reducing expressions until they cannot be simplified any further. This is essential for both making sense of functions and solving equations efficiently.

In the context of function composition, simplification often plays a big role. For example, when we focus on our exercise:

• Calculating \((f \circ g)(x)\) leads us to an expression \(1 - (1 - x)\), which simplifies neatly to \(x\).
• Another simplification occurs with \((g \circ f)(x)\), where \( \frac{5x}{5} \) reduces to \(x\).

In each case, simplification helps us see that the composed functions are in fact linearly simple, often revealing properties about how they transform inputs. Mastering simplification helps you not only solve math problems effectively but also understand the behavior and limitations of equations and models you encounter across different areas of math.