Problem 62 - 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. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=\sqrt{x}, \quad g(x)=1-x^{2} $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = \sqrt{1 - x^2}\), domain: \([-1, 1]\); (b) \((g \circ f)(x) = 1 - x\), domain: \([0, \infty)\); (c) \((f \circ f)(x) = \sqrt[4]{x}\), domain: \([0, \infty)\).

Step 1: Find the Composite Function (f ∘ g)(x)

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). Since \(f(x) = \sqrt{x}\), substitute \(1 - x^2\) for \(x\) in \(f\), giving us \(f(g(x)) = \sqrt{1 - x^2}\). Thus, \((f \circ g)(x) = \sqrt{1 - x^2}\).

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

The domain of \((f \circ g)(x)\) is where \(1 - x^2\) is non-negative, because the square root function requires non-negative inputs. Therefore, solve \(1 - x^2 \geq 0\): \(-1 \leq x \leq 1\). Thus, the domain of \((f \circ g)(x)\) is \([-1, 1]\).

Step 3: Find the Composite Function (g ∘ f)(x)

For \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). Here, \(g(x) = 1 - x^2\). Substitute \(\sqrt{x}\) for \(x\): \(g(f(x)) = 1 - (\sqrt{x})^2 = 1 - x\). Therefore, \((g \circ f)(x) = 1 - x\).

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

The function \((g \circ f)(x) = 1 - x\) has a domain that needs to consider the domain of \(f(x)\), which is \(x \geq 0\). Thus, the domain of \((g \circ f)(x)\) is \([0, \infty)\).

Step 5: Find the Function (f ∘ f)(x)

To find \((f \circ f)(x)\), substitute \(f(x)\) into itself. Since \(f(x) = \sqrt{x}\), \(f(f(x)) = \sqrt{\sqrt{x}} = \sqrt[4]{x}\). Thus, \((f \circ f)(x) = \sqrt[4]{x}\).

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

The function \(\sqrt[4]{x}\) requires that \(x \geq 0\) because the square root is defined for non-negative numbers. Therefore, the domain of \((f \circ f)(x)\) is \([0, \infty)\).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial to work effectively with mathematical expressions. The domain of a function refers to all the possible input values (often "x" values) that the function can accept without leading to undefined behavior or breaking any mathematical rules.
For instance, consider the square root function, which is defined as \( f(x) = \sqrt{x} \). The requirement for this function is that the input \( x \) must be a non-negative number (i.e., \( x \geq 0 \)) because the square root of a negative number is not defined in the real number system.

• For the composite function \((f \circ g)(x) = \sqrt{1 - x^2}\), the domain is determined where \( 1 - x^2 \) is non-negative, being that square root requires non-negative input values.
• For \((g \circ f)(x) = 1 - x\), ensure that the domain of \(f(x)\) is respected, thus it applies to \([0, \infty)\).
• Similarly, \((f \circ f)(x) = \sqrt[4]{x}\) can only be computed for non-negative \(x\), so the domain is \([0, \infty)\).

Understanding domains helps set constraints, allowing for correctly working with functions and their compositions.

Composite Functions

Composite functions are created by combining two or more functions. The process involves taking an output from one function and using it as an input for another function. In symbols, for two functions \( f(x) \) and \( g(x) \), their composite is denoted as \((f \circ g)(x)\), which is read as "f of g of x."
To form \((f \circ g)(x)\), you plug the expression for \(g(x)\) directly into the expression for \(f(x)\). Practically, this requires replacing the \(x\) in \(f(x)\) with \(g(x)\). In our example, \( g(x) = 1 - x^2 \) is substituted into \( f(x) = \sqrt{x} \), resulting in \( \sqrt{1 - x^2} \), which is the composite function.

• Composite functions allow for understanding how multiple independent effects are chained together, valuable for complex problem-solving scenarios.
• Always ensure the domain of the composite respects the domain of the individual functions, as it restricts valid inputs.

Through practice, working with composites becomes more intuitive, facilitating deeper insights into mathematical behavior.

Square Root Functions

Square root functions, like other radical functions, have unique properties worth understanding to solve accurately. These functions are primarily represented as \( f(x) = \sqrt{x} \) and can extend in forms such as \( \sqrt{1-x^2} \) or \( \sqrt[4]{x} \).
The graph of a basic square root function \( f(x) = \sqrt{x} \) begins at the origin \((0,0)\) and extends to the right, curving upwards. This indicates that the function only accepts non-negative inputs due to the square root's inability to handle negative numbers without producing imaginary results.

• The expression \( \sqrt{x} \) must have \( x \geq 0 \).
• For functions like \( \sqrt{1-x^2} \), the condition \( 1-x^2 \geq 0 \) must be met, limiting the domain to values of \(x\) that result in non-negative square roots.
• Multi-layered square root expressions, such as \( \sqrt[4]{x} \), further emphasize the requirement for non-negative \(x\).

Understanding these restrictions ensures the correct evaluation and graphing of square root functions across growth and transformation contexts.