Problem 25 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\sqrt{x-2}, \quad g(x)=\sqrt{x+5} $$

Step-by-Step Solution

Answer

\((f \circ g)(x) = \sqrt{\sqrt{x+5} - 2}, x \geq -1; (g \circ f)(x) = \sqrt{\sqrt{x-2} + 5}, x \geq 2.\)

Step 1: Understand the functions

We have two functions, \( f(x) = \sqrt{x - 2} \) and \( g(x) = \sqrt{x + 5} \). The function \( f(x) \) is defined for \( x \geq 2 \) because the expression under the square root must be non-negative. Similarly, \( g(x) \) requires \( x \geq -5 \).

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

The composition \((f \circ g)(x)\) means applying \( g(x) \) first and then \( f \) on the result. Substitute \( g(x) = \sqrt{x+5} \) into \( f(x) \):\[ f(g(x)) = f(\sqrt{x+5}) = \sqrt{\sqrt{x+5} - 2} \]

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

The domain of \( f \circ g \) requires determining when each part is defined. First, \( g(x) = \sqrt{x+5} \) requires \( x \geq -5 \). Then, for \( f(y) = \sqrt{y - 2} \), we need \( \sqrt{x+5} \geq 2 \), which simplifies to \( x+5 \geq 4 \) or \( x \geq -1 \). Thus, the domain is \( x \geq -1 \).

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

For the composition \((g \circ f)(x)\), apply \( f(x) \) first then \( g \). Substitute \( f(x) = \sqrt{x-2} \) into \( g(x) \):\[ g(f(x)) = g(\sqrt{x-2}) = \sqrt{\sqrt{x-2} + 5} \]

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

Start with \( f(x) = \sqrt{x-2} \), which requires \( x \geq 2 \). Then, solve for \( g(y) = \sqrt{y + 5} \) defined at \( y \geq -5 \). However, since \( y = \sqrt{x-2} \), which is always positive or zero for \( x \geq 2 \), this condition simplifies to \( x \geq 2 \). Thus, the domain is \( x \geq 2 \).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial in determining where the function is valid or defined. Simply put, the domain is the set of all possible input values (usually represented as \( x \)) that allow the function to work without issues like division by zero or taking the square root of a negative number. In this exercise, we look at two functions: \( f(x) = \sqrt{x - 2} \) and \( g(x) = \sqrt{x + 5} \).

• For \( f(x) = \sqrt{x - 2} \), the domain is \( x \geq 2 \) because you cannot take the square root of a negative value.
• On the other hand, for \( g(x) = \sqrt{x + 5} \), the domain is \( x \geq -5 \).

When dealing with composite functions, understanding the domain becomes a bit more complex, as you have to consider the restrictions of each individual function involved.

Square Root Function

A square root function is a type of function that involves the square root of an expression. The expression inside the square root symbol must be non-negative for the function to have real values. This is because real number square roots of negative numbers are not defined within the real number system.In our example, we have two square root functions: \( f(x) = \sqrt{x - 2} \) and \( g(x) = \sqrt{x + 5} \).

• In \( f(x) \), the part \( x - 2 \) must be greater than or equal to zero, leading to the requirement \( x \geq 2 \).
• For \( g(x) \), the part \( x + 5 \) must also be non-negative, resulting in \( x \geq -5 \).

Square root functions often define the domain of a composite function because they contain inherent restrictions related to negative values.

Composite Functions

Composite functions are functions that apply one function to the results of another. You might see them written as \((f \circ g)(x)\), which reads as "f of g of x." This means you are applying \( g(x) \) first, then applying \( f \) to that result.In our exercise, we are instructed to find both \((f \circ g)(x)\) and \((g \circ f)(x)\).

• To find \((f \circ g)(x)\), substitute \( g(x) = \sqrt{x + 5} \) into \( f(x) \), resulting in \( \sqrt{\sqrt{x+5} - 2} \).
• For \((g \circ f)(x)\), substitute \( f(x) = \sqrt{x-2} \) into \( g(x) \), giving you \( \sqrt{\sqrt{x-2} + 5} \).

The key to solving composite functions is careful substitution and then determining any new domain restrictions that may arise. For each composite setup, you need to ensure both inner and resulting operations are defined.

Function Notation

Function notation provides a concise way to display the operations and processes of a function. It's essential in expressing and solving composite functions. For instance, \( f(x)\) and \( g(x)\) mean we apply certain rules or operations to the variable \( x \).

• \( f(x) = \sqrt{x - 2} \) indicates that for every input \( x \), the result is the square root of \( x - 2 \).
• Similarly, \( g(x) = \sqrt{x + 5} \) means each input \( x \) yields the square root of \( x + 5 \).

In composite function notation, such as \((f \circ g)(x)\), we interpret this as first computing \( g(x) \) and then applying \( f \) to the result. Function notation simplifies writing and solving expressions that involve multiple functions and clarifies the order of operations.