Problem 71 - 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. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=\frac{1}{k x}, k>0, \quad g(x)=\frac{1}{k x}, k>0 $$

Step-by-Step Solution

Answer

(f ∘ g)(x) = x, (g ∘ f)(x) = x, (f ∘ f)(x) = x; domain for all is ℝ\{0}.

Step 1: Understand Composite Functions

The notation \((f \circ g)(x)\) means \(f(g(x))\), i.e., substitute \(g(x)\) into \(f(x)\). Similarly, \((g \circ f)(x)\) means \(g(f(x))\), and \((f \circ f)(x)\) means \(f(f(x))\).

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

First, substitute the function \(g(x) = \frac{1}{kx}\) into \(f(x)\). Thus, \(f(g(x)) = f\left(\frac{1}{kx}\right) = \frac{1}{k \left(\frac{1}{kx}\right)} = x\). Therefore, \((f \circ g)(x) = x\).

Step 3: Determine the Domain of \((f \circ g)(x)\)

Since \((f \circ g)(x) = x\) is a linear function, its domain is all real numbers, i.e., \(\mathbb{R}\). However, given the form of \(g(x)\) and the constraint \(kx eq 0\), \(x eq 0\), the domain of \((f \circ g)(x)\) is all real numbers except \(x = 0\), which is \(\mathbb{R} \setminus \{0\}\).

Step 4: Calculate \((g \circ f)(x)\)

Substitute \(f(x) = \frac{1}{kx}\) into \(g(x)\). Thus, \(g(f(x)) = g\left(\frac{1}{kx}\right) = \frac{1}{k \left(\frac{1}{kx}\right)} = x\). Therefore, \((g \circ f)(x) = x\).

Step 5: Determine the Domain of \((g \circ f)(x)\)

Similar to step 3, \((g \circ f)(x) = x\) is a linear function, and its domain would be all real numbers. However, \(f(x) = \frac{1}{kx}\) requires \(x eq 0\), so the domain of \((g \circ f)(x)\) is \(\mathbb{R} \setminus \{0\}\).

Step 6: Calculate \((f \circ f)(x)\)

To find \((f \circ f)(x)\), substitute \(f(x) = \frac{1}{kx}\) into itself. Thus, \(f(f(x)) = f\left(\frac{1}{kx}\right) = \frac{1}{k \left(\frac{1}{kx}\right)} = x\). Therefore, \((f \circ f)(x) = x\).

Step 7: Determine the Domain of \((f \circ f)(x)\)

Since we substitute \(f(x)\) into itself, \(x\) must satisfy \(kx eq 0\). This implies \(x eq 0\). So, the domain of \((f \circ f)(x)\) is \(\mathbb{R} \setminus \{0\}\).

Key Concepts

Function Composition

Function composition is a way to combine two functions into a single function. It involves taking the output of one function and using it as the input of another. So, if you have two functions, say \(f(x)\) and \(g(x)\), you can create new functions known as composed functions. The process is usually denoted with the notation \((f \circ g)(x)\), which means \(f(g(x))\). This notation implies that you take the function \(g(x)\), calculate its value, and then substitute that result into the function \(f(x)\).
For example, if \(f(x) = \frac{1}{kx}\) and \(g(x) = \frac{1}{kx}\), to find \((f \circ g)(x)\), you substitute \(g(x)\) into \(f(x)\), which ends up simplifying to just \(x\). Composed functions are useful in various mathematics and applied fields since they allow for more complex functions to be built from simpler ones.
When working with function composition, it's important to understand the operation order: calculate within the inner function first, and then use that result in the outer function.

Domain of Functions

The domain of a function is the set of all input values \(x\) that the function can accept without causing any mathematical errors, like division by zero or taking a square root of a negative number in real numbers.
For the function \(f(x) = \frac{1}{kx}\), the primary concern in the domain is ensuring we do not divide by zero. In this case, \(kx eq 0\) means \(x eq 0\). So, the domain of \(f(x)\) would be all real numbers except zero, denoted as \(\mathbb{R} \setminus \{0\}\).
When you are dealing with composed functions like \((f \circ g)(x)\), you must consider the domains of both individual functions. You need to find where both \(g(x)\) and the composition \(f(g(x))\) are defined without any restrictions. The domain is effectively the intersection of these subsets to ensure the composed function makes sense at the analyzed points.

Rational Functions

Rational functions are functions of the form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) is not zero. The characteristic feature of these functions is that they can have undefined points, known as asymptotes, typically where the denominator equals zero.
For the rational function \(f(x) = \frac{1}{kx}\), even though it looks simple, we must be careful about the denominator. If \(x = 0\), the function is undefined since division by zero is not permissible in mathematics. As a result, although rational functions can often be simplified, their domain restriction on division by zero cannot be ignored.
Graphically, rational functions can present vertical asymptotes at the roots of the denominator, holes where terms are canceled out, and horizontal or oblique asymptotes if the degrees of the polynomials meet specific conditions. Studying these can often give a good insight into the behavior of rational functions.