Problem 25 - The Calculus with Analytic Geometry

Question

Functions \(f\) and \(g\) are defined. In each exercise define \(f \circ g\), and determine all values of \(x\) for which \(f \circ g\) is continuous. $$ f(x)=\frac{1}{x-2} ; g(x)=\sqrt{x} $$

Step-by-Step Solution

Answer

The function \( f \circ g(x) = \frac{1}{\sqrt{x} - 2} \) is continuous for all \( x \geq 0 \) except \( x = 4 \).

Step 1: Understand the Functions

Analyze the given functions. The function \( f(x) = \frac{1}{x-2} \) is defined for all \( x \) except \( x = 2 \). The function \( g(x) = \sqrt{x} \) is defined for all \( x \geq 0 \).

Step 2: Define the Composition

The composition \( (f \circ g)(x) \) means applying \( g(x) \) first and then \( f(x) \). So, \( f \circ g(x) = f(g(x)) \). Substitute \( g(x) \) into \( f(x) \).

Step 3: Substitute and Simplify

Substitute \( g(x) = \sqrt{x} \) into \( f(x) \). Therefore, \( f(g(x)) = f(\sqrt{x}) = \frac{1}{\sqrt{x} - 2} \).

Step 4: Determine the Domain for Continuity

For \( f \circ g(x) \) to be continuous, \( \sqrt{x} - 2 eq 0 \) and \( \sqrt{x} \geq 0 \). Set \( \sqrt{x} - 2 eq 0 \) which implies \( \sqrt{x} eq 2 \). Squaring both sides, \( x eq 4 \). Also, \( \sqrt{x} \geq 0 \) means \( x \geq 0 \). Therefore, \( x \) must be \( x \geq 0 \) and \( x eq 4 \).

Key Concepts

Continuity

Understanding the continuity of a function is crucial in mathematics. A function is continuous at a point if there are no breaks, holes, or jumps at that point. More formally, a function \( f(x) \) is continuous at \( x=a \) if the following three conditions are met:
1. \( f(a) \) is defined.
2. The limit \( \displaystyle\lim_{x \to\ a} f(x) \) exists.
3. The limit of \( f(x) \) as \( x \) approaches \( a \) equals \( f(a) \): \( \displaystyle\lim_{x \to\ a} f(x) = f(a) \).
In the context of the composite function \( f \circ g(x) \), it is essential to check for continuity at every point within the domain of \( g(x) \). Simply put, this involves ensuring that:\( f(g(x)) \) is defined for all \( x \) in the considered interval, and exploring any potential discontinuities.
To confirm continuity, we should carefully evaluate the combined behavior of each function at critical points within the specified range.

Domain of Functions

The domain of a function refers to the set of possible input values \( x \) for which the function is defined. For example, the domain of \( f(x) = \frac{1}{x-2} \) excludes \( x = 2 \) because division by zero is undefined. Similarly, \( g(x) = \sqrt{x} \) is only defined for \( x \geq 0 \).
When dealing with composite functions, such as \( f \circ g \), determining the domain involves finding the common set of values where both functions are valid. This requires:
- Evaluating where each individual function is defined.
- Identifying overlaps where both constraints are satisfied.
In our example, \( f \circ g(x) = \frac{1}{\sqrt{x} - 2} \) implies that \( \sqrt{x} \) must be greater than 2, leading to \( x e 4 \) and \( x \geq 0 \). Thus, the domain is \( x \geq 0 \) except \( x e 4 \).

Composite Function

A composite function is created when one function is applied to the result of another function. This is denoted as \( (f \circ g)(x) = f(g(x)) \). Essentially, you first apply \( g(x) \) to \( x \), and then apply \( f \) to the result of \( g(x) \).
For instance, given \( f(x) = \frac{1}{x-2} \) and \( g(x) = \sqrt{x} \), the composition \( f \circ g(x) \) involves plugging \( g(x) \) into \( f \). Thus, \( f(g(x)) = \frac{1}{\sqrt{x} - 2} \).
When finding the composite function, pay attention to how the domain of \( g(x) \) influences the domain of \( f \). In our case, \( g(x) \) produces values \( \sqrt{x} \geq 0 \), and the expression \( \frac{1}{\sqrt{x} - 2} \) must be defined, avoiding division by zero. Hence, the composite function \( f \circ g \) is valid over a specific range of \( x \) values.