Problem 67 - 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. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=\sqrt{x-1}, \quad g(x)=3 x $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = \sqrt{3x - 1}\), domain \([\frac{1}{3}, \infty)\); (b) \((g \circ f)(x) = 3\sqrt{x-1}\), domain \([1, \infty)\); (c) \((f \circ f)(x) = \sqrt{\sqrt{x-1} - 1}\), domain \([2, \infty)\).

Step 1: Determine (f ∘ g)(x)

The composition \((f \circ g)(x)\) means substituting \(g(x)\) into \(f(x)\). Since \(f(x) = \sqrt{x-1}\) and \(g(x) = 3x\), we find \((f \circ g)(x) = f(g(x)) = \sqrt{3x - 1}\).

Step 2: Determine Domain of (f ∘ g)(x)

For \((f \circ g)(x) = \sqrt{3x - 1}\) to be defined, the expression inside the square root must be non-negative. Solve \(3x - 1 \geq 0\) to get \(x \geq \frac{1}{3}\). Therefore, the domain is \(\left[\frac{1}{3}, \infty\right)\).

Step 3: Determine (g ∘ f)(x)

The composition \((g \circ f)(x)\) means substituting \(f(x)\) into \(g(x)\). Since \(g(x) = 3x\) and \(f(x) = \sqrt{x-1}\), we have \((g \circ f)(x) = g(f(x)) = 3(\sqrt{x-1})\).

Step 4: Determine Domain of (g ∘ f)(x)

For \((g \circ f)(x) = 3(\sqrt{x-1})\) to be defined, \(x-1\) must be non-negative. Solve \(x-1 \geq 0\) to get \(x \geq 1\). Therefore, the domain is \([1, \infty)\).

Step 5: Determine (f ∘ f)(x)

The composition \((f \circ f)(x)\) means substituting \(f(x)\) into itself. \(f(x) = \sqrt{x-1}\), so \((f \circ f)(x) = f(f(x)) = \sqrt{\sqrt{x-1} - 1}\).

Step 6: Determine Domain of (f ∘ f)(x)

For \((f \circ f)(x) = \sqrt{\sqrt{x-1} - 1}\) to be defined, \(\sqrt{x-1} - 1\) must be non-negative. Solve \(\sqrt{x-1} \geq 1\) to get \(x-1 \geq 1\), which implies \(x \geq 2\). Therefore, the domain is \([2, \infty)\).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (often seen as "x" values) for which the function is defined. Essentially, it's the collection of numbers that you can safely plug into the function without causing any mathematical mishaps. For instance, you can't include numbers that result in dividing by zero or taking the square root of a negative number, as these actions are undefined in the set of real numbers.

To determine the domain, ask yourself: Are there any values that will cause problems if plugged into the function? If the answer is yes, then those values are excluded from the domain.

Steps to find out the domain:

• Look for any operations that restrict inputs, like square roots or divisions.
• Address any variables inside these operations, ensuring conditions like non-negativity for square roots are satisfied.
• Write the domain as an interval or union of intervals that highlight all valid inputs.

In the original problem, determining the domain of each composite function was crucial to ensure the function is well-defined.

Composite Functions

Composite functions occur when you combine two functions, say \( f(x) \) and \( g(x) \), into one. The combination is denoted as \( (f \circ g)(x) \), which reads as "f of g of x." This means you first apply \( g(x) \) and then use that result as the input for \( f(x) \).

Steps involved in creating composite functions:

• Identify the two functions you are working with.
• Substitute the inside function (\( g(x) \)) into the outer function (\( f(x) \)).

The order of composition matters; \( (f \circ g)(x) \) is not the same as \( (g \circ f)(x) \), because you apply the functions in different sequences.

For the original exercise, this involved:

• Calculating \( (f \circ g)(x) = \sqrt{3x - 1} \).
• Calculating \( (g \circ f)(x) = 3(\sqrt{x-1}) \).
• Calculating \( (f \circ f)(x) = \sqrt{\sqrt{x-1} - 1} \).

Understanding function composition is key to correctly solving and determining domains of each function in combination.

Square Root Function

A square root function is a type of function that involves the square root of a variable. The basic form is \( f(x) = \sqrt{x} \), where \( x \) must be a non-negative number, because square roots of negative numbers are not defined in the realm of real numbers.

Characteristics of the square root function:

• The output values will always be non-negative, as we can't have negative square roots in real numbers.
• The function increases as \( x \) increases, but at a decreasing rate, meaning it rises more slowly as x gets larger.
• The domain of \( f(x) = \sqrt{x} \) is \([0, \infty)\), as all non-negative numbers can be input.

In composite functions like \( (f \circ g)(x) = \sqrt{3x - 1} \), the expression \( 3x - 1 \) must be non-negative. Solving \( 3x - 1 \geq 0 \), the domain found was \([\frac{1}{3}, \infty)\). Similarly, when \( (f \circ f)(x) = \sqrt{\sqrt{x-1} - 1} \), further conditions are applied to maintain non-negative conditions at each nested square root step.