Problem 21 - 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. $$ f(x)=x^{3} ; g(x)=\sqrt{x} $$

Step-by-Step Solution

Answer

The composite function is \(x^{3/2}\) and it is continuous for \(x \geq 0\).

Step 1: Define the Composition of Functions

To find the composition of the functions, substitute the function g(x) into the function f(x). Thus, \[ (f \circ g)(x) = f(g(x)) \] Substituting: \[ (f \circ g)(x) = f(\sqrt{x}) \] Since \(f(x) = x^3\), we replace \(x\) with \(\sqrt{x}\): \[ (f \circ g)(x) = \left(\sqrt{x}\right)^3 = x^{3/2} \]

Step 2: Determine the Continuity of the Composite Function

We need to determine the values of \(x\) for which the function \(x^{3/2}\) is continuous. Since \(x^{3/2}\) is defined for all non-negative values of \(x\) (because we cannot take the square root of a negative number), the domain of \(x^{3/2}\) is \(x \geq 0\). Polynomial functions like \(x^{3/2}\) are continuous over their domain.

Key Concepts

Composition of Functions

When we talk about the composition of functions, we are essentially combining two functions to produce a new one. Here's how it works: if you have two functions, say \( f(x) \) and \( g(x) \), the composition \( f \circ g \) means applying \( g(x) \) first and then applying \( f \) to the result from \( g(x) \). Mathematically, it is represented as \( (f \circ g)(x) = f(g(x)) \).

Let's break this down with our original functions:

\[ f(x) = x^{3}, \, g(x) = \, \sqrt{x} \]

To find \( (f \circ g)(x) \), we need to plug \( g(x) \) into \( f(x) \):
\[ (f \circ g)(x) = f(g(x)) \ = f(\sqrt{x}) \]
Then, we substitute \( \sqrt{x} \) into \( f(x) \):
\[ f(\sqrt{x}) = (\sqrt{x})^3 = x^{3/2} \]

This new function \( x^{3/2} \) is the composition of the functions \( f \) and \( g \). This shows why composition of functions is such a powerful concept. It lets us combine different functions into one, providing a new perspective in problem-solving.

Continuity of Functions

Continuity is an essential concept in calculus and mathematics in general. A function is considered continuous at a point if there is no interruption in the graph of the function at that point. More formally, a function \( f(x) \) is continuous at point \( c \) if the following three conditions are satisfied:

• The function is defined at \( c \); \( f(c) \) exists.
• The limit \( \lim_{{x \to c}} f(x) \) exists.
• The limit and the function value are equal; \( \lim_{{x \to c}} f(x) = f(c) \).

Polynomial functions, as in our example, are usually continuous over their entire domains. For the composite function \( (f \circ g)(x) = x^{3/2} \), we need to determine where it is continuous. The function \( x^{3/2} \) is continuous wherever it is defined since it follows the properties of polynomial functions.

Domain of Composite Functions

Determining the domain of composite functions is crucial for understanding where the function is valid and continuous. The domain of a composite function \( (f \circ g)(x) \) is influenced by the domains of both \( f(x) \) and \( g(x) \). Specifically, it involves the following two criteria:

• The values for which \( g(x) \) is defined.
• The values for which \( f(g(x)) \) is defined.

Our original functions are:

• \( f(x) = x^3 \), defined for all real \( x \).
• \( g(x) = \sqrt{x} \), defined for \( x \geq 0 \).

Therefore, the domain of \( g(x) \) is from \( x = 0 \) to \( x = \infty \), and we need \( g(x) \) to be defined to use it in \( f(g(x)) \). Consequently, the composite function \( (f \circ g)(x) = x^{3/2} \) is defined and continuous for \( x \geq 0 \). This intersection of domains tells us where the composite function is valid. By considering these steps, you can handle the domain of any composite function with confidence.