Problem 32 - College Algebra

Question

Find: a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=\frac{x}{x+5}, g(x)=\frac{6}{x}$$

Step-by-Step Solution

Answer

\(f(g(x)) = \frac{6}{6 + 5x}\) with the domain being all real numbers except \(x = -\frac{6}{5}\) and \(x = 0\).

Step 1: Find (f ∘ g)(x)

Substitute \(g(x)\) into \(f(x)\). To find \(f(g(x))\), place \(g(x) = \frac{6}{x}\) into \(f(x)\) wherever \(x\) occurs. This gives \(f(g(x)) = \frac{\frac{6}{x}}{\frac{6}{x} + 5}\).

Step 2: Simplify (f ∘ g)(x)

Simplify that complex fraction above to obtain a simpler form. Multiply the numerator and the denominator of the main fraction by \(x\) to eliminate the fraction in the denominator, then simplify. Yielding, \(f(g(x)) = \frac{6}{6 + 5x}\).

Step 3: Determine domain of (f ∘ g)(x)

The denominator of the fraction cannot be equal to zero as it will make the whole function undefined. So, set \(6 + 5x ≠ 0\). Solving for \(x\) gives \(x ≠ -\frac{6}{5}\). Furthermore, \(g(x) = \frac{6}{x}\) is not defined for \(x = 0\). So, togetherv, the domain of \(f ∘ g\) is all real numbers except \(x = -\frac{6}{5}\) and \(x = 0\)

Key Concepts

Domain of a Function

Understanding the domain of a function is like knowing the limits within which a function can operate. It's the set of all possible inputs, or 'x' values, that can be plugged into the function without causing any mathematical inconsistencies, such as division by zero.

Let's take our composite function \( f \( g(x) \) \). As we've seen in the step-by-step solution, to find the domain, we check for the conditions that make the function's denominator zero, as these are not allowed. For the composite function, this is where \( 6 + 5x \) is equal to zero. Additionally, since \( g(x)=\frac{6}{x} \) is part of our composite, \( x \) cannot equal to zero either, or we would have a division by zero. This is why the domain of \( f \( g(x) \) \) excludes \( x = -\frac{6}{5} \) and \( x = 0 \).

Always remember, when working with compositions, we must consider the domains of both functions involved.

Function Composition

Composition of functions is a little like a cooking recipe: we combine ingredients in a specific order to create a new dish. In mathematics, we combine functions. When we write \( f \( g(x) \) \), it means we first apply the function \( g \) to the input \( x \) and then take the result of that and use it as the input to the function \( f \).

The exercise you worked on is a classic case of function composition where we placed \( g(x)=\frac{6}{x} \) into \( f(x) \) for every instance of \( x \) and simplified to find the new composite function. One key takeaway is that the order in which functions are composed matters – \( f \( g(x) \) \) is generally not the same as \( g \( f(x) \) \). This is similar to understanding that the steps in a recipe can't be rearranged randomly if you want your dish to come out correctly.

Simplifying Complex Fractions

Complex fractions can look intimidating with fractions piled upon other fractions. Simplifying them makes them much more manageable. Our composite function resulted in a complex fraction.

To simplify, as shown in the step-by-step solution, one approach is to multiply the top and bottom of the main fraction by the least common denominator (LCD) of the fractions within our complex fraction. In our case, that's \( x \), the denominator of the inner fraction \( g(x) \). This clears out the 'fraction within a fraction', leaving us with \( f \( g(x) \) = \frac{6}{6 + 5x} \), a much simpler expression to work with.

Understanding how to simplify complex fractions is powerful. It not only eases calculations but can also reveal characteristics of the function that may not have been immediately evident.

Algebraic Functions

Algebraic functions are the bread and butter of high school math. They are functions that involve only algebraic operations, such as addition, subtraction, multiplication, division, and raising to a power. In our exercise, both \( f(x) = \frac{x}{x+5} \) and \( g(x) = \frac{6}{x} \) are algebraic functions because they're built from these basic operations.

When you work with algebraic functions, you have a whole toolkit of algebraic manipulations at your disposal to address the task at hand. This includes simplifying expressions, factoring, finding common denominators, and canceling terms. By mastering these skills, you can tackle a wide range of problems, such as finding the domain or composing functions like we did in this exercise.