Problem 70 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

a. The composite function \(f \circ g(x) = \frac{6}{5x+6}\)\nb. The domain of the function \(f \circ g\) is all real numbers \(x\) such that \(x \neq -\frac{6}{5}\)

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

First, replace every \(x\) in function \(f(x)\) with \(g(x)\) to get: \(f(g(x)) = \frac{g(x)}{g(x)+5}\). Substituting the given \(g(x) = \frac{6}{x}\) into the expression, it will read: \(f \circ g(x) = \frac{\frac{6}{x}}{\frac{6}{x} + 5}\).

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

To simplify this complex rational expression, multiply the numerator and the denominator of the fraction by \(x\) to get rid of the fractions in the denominator which gives: \(f \circ g(x) = \frac{6}{\frac{6}{x} \cdot x + 5x} = \frac{6}{6 + 5x} = \frac{6}{5x+6}\). So, \(f \circ g(x) = \frac{6}{5x+6}\).

Step 3: Finding the domain of \(f \circ g\)

In this step, it is required to find the set of all real x-values for which the function \(f \circ g\) gives a real output. The function \(f \circ g(x) = \frac{6}{5x+6}\) will be undefined when its denominator equals zero. Therefore, the domain of \(f \circ g(x)\) will be all real numbers except the value of \(x\) that makes \(5x+6 = 0\). Solving for \(x\) gives \(x = -\frac{6}{5}\). Therefore, the domain of \(f \circ g(x)\) is \(x \neq -\frac{6}{5}\).

Key Concepts

Domain of a Function

The domain of a function is all about figuring out which values can be plugged into a function without breaking any mathematical rules. When you're dealing with functions, it's like determining the range of possible inputs where the function behaves nicely and produces real outputs. It’s important to determine the domain before proceeding with other operations on functions.
For instance, in the function given in the exercise, the expression for the composition of functions is \[f \circ g(x) = \frac{6}{5x+6}\].

We see that fractions cannot have zero in the denominator because division by zero is undefined. Therefore, the domain consists of all real numbers except the value that makes the denominator zero. In this case, solving the equation \(5x + 6 = 0\) reveals \(x = -\frac{6}{5}\).

So, the domain of \(f \circ g(x)\) would be all real numbers except \(x = -\frac{6}{5}\). Always watch out for these restrictions on the denominator when dealing with rational expressions.

Rational Expressions

Rational expressions are like fractions, but instead of just numbers in the numerator and denominator, they have variables. They can sometimes seem tricky, but breaking them down step-by-step is the key.
In the provided exercise, the function \(g(x)\) is a rational expression given by \(\frac{6}{x}\). To find \(f \circ g(x)\), we substituted \(g(x)\) into \(f(x)\), yielding \(\frac{\frac{6}{x}}{\frac{6}{x} + 5}\).

This expression is complex but can be simplified by eliminating the fraction in a fraction. By multiplying both numerator and denominator by \(x\), we simplify to \(\frac{6}{5x+6}\).

Manipulating rational expressions often involves these steps:

• Simplify the expression by finding a common denominator.
• Eliminate complex fractions by multiplying by a suitable expression.
• Factor and reduce the expression wherever possible.

Pay attention to each step to avoid errors and ensure the expression is in its simplest form.

Inverse Functions

Inverse functions are about reversing a process. If you have a function that turns input \(a\) into output \(b\), then the inverse turns \(b\) back into \(a\). It's like having a lock and a key – each function unlocks the output to go back to the original input.
To find an inverse, we typically need to swap the roles of the input and output and then solve for the new input. So, for a function \(f(x)\), its inverse, denoted as \(f^{-1}(x)\), satisfies \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).

However, not all functions have inverses. For a function to have an inverse, it must be bijective, meaning it's both one-to-one (every element of the range is mapped by exactly one element of the domain) and onto (every element of the codomain is covered).

In the context of rational expressions, it's crucial to determine which functions have inverses before attempting to find them. Start by checking if the function is bijective. If it meets this criterion, you're set to find its inverse by swapping \(x\) and \(y\) (or inputs and outputs) and solving for the input.