Problem 28 - 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. $$ f(x)=\sqrt{x+1} ; g(x)=\sqrt[3]{x} $$

Step-by-Step Solution

Answer

The function \( f \circ g(x) = \sqrt{\sqrt[3]{x} + 1} \) is continuous for all \( x \geq -1 \).

Step 1: Understand the Definitions

First, identify the given functions: \( f(x) = \sqrt{x+1} \) and \( g(x) = \sqrt[3]{x} \).

Step 2: Compose the Functions

Find \( f \circ g \), which means \( f(g(x)) \). Substitute \( g(x) \) into \( f \).

Step 3: Substitute and Simplify

Substitute \( g(x) = \sqrt[3]{x} \) into \( f(x) \): \( f(g(x)) = f(\sqrt[3]{x}) = \sqrt{\sqrt[3]{x} + 1} \). So, \( f \circ g(x) = \sqrt{\sqrt[3]{x} + 1} \).

Step 4: Determine the Domain

The function \( f(x) = \sqrt{x+1} \) is defined for \( x+1 \geq 0 \), or \( x \geq -1 \). Applying this to \( \sqrt[3]{x} + 1 \geq 0 \), solve for \( x \): \( \sqrt[3]{x} \geq -1 \). Since the cube root function is continuous and its output can take any real number value, find the \( x \) values for which \( \sqrt[3]{x} \geq -1 \). Hence, \( x \geq -1 \).

Step 5: Conclusion

The function \( f \circ g(x) = \sqrt{\sqrt[3]{x} + 1} \) is continuous for all \( x \geq -1 \). Thus, the composition is defined and continuous for these values of \( x \).

Key Concepts

Domain of Functions

Understanding the domain of a function means identifying all possible input values (x) for which the function produces a valid output.
For the function \( f(x) = \sqrt{x+1} \), we need to ensure the expression inside the square root is non-negative.
This means solving \( x+1 \geq 0 \) or \( x \geq -1 \).
For the cube root function \( g(x) = \sqrt[3]{x} \), the domain is all real numbers, since the cube root of any real number is always a real number itself.
When we compose functions, like finding \( f(g(x)) \), we must consider the domains of both functions involved. The composite function \( f \circ g(x) = f(g(x)) \), which becomes \( \sqrt{\sqrt[3]{x} + 1} \), will have the domain determined by the inner function \( g(x) \) and the resulting expression inside \( f(x) \).
This leads us to solve for \( \sqrt[3]{x} + 1 \geq 0 \), giving \( x \geq -1 \) as the domain for the composite function.

Continuity

Continuity of a function at a point means there is no interruption or jump in its graph at that point.
For a function to be continuous, the limit approaching that point from both sides must equal the function's value at that point.
The function \( f(x) = \sqrt{x+1} \) is continuous for all \( x \geq -1 \) because the square root function is continuous for non-negative inputs.
The function \( g(x) = \sqrt[3]{x} \) is continuous for all real numbers since the cube root function is inherently continuous everywhere in its domain.
When composing these functions, we see the composite function \( f \circ g(x) = \sqrt{\sqrt[3]{x} + 1} \) is continuous as well for all \( x \geq -1 \), as both functions involved are continuous at these points.
This means there is no break, hole, or jump in the graph of the composite function \( f \circ g \) within this domain.

Composite Functions

Composite functions are formed when you apply one function to the results of another.
The notation \( f \circ g \) represents the composition of functions \( f \) and \( g \), meaning \( f(g(x)) \).
In our example, \( f(x) \) is applied to \( g(x) \), so we substitute \( g(x) = \sqrt[3]{x} \) into \( f \): \( f(g(x)) = f(\sqrt[3]{x}) \).
This results in \( f(g(x)) = \sqrt{\sqrt[3]{x} + 1} \), giving us the composite function \( f \circ g(x) \).
It’s essential to determine the domain of this composite function by considering the valid input values for both functions and ensuring they are non-negative within the context of square roots.
Composite functions can sometimes be complex, but understanding the domains and continuity of individual functions helps simplify the process.