Problem 102 - Calculus: Early Transcendentals

Question

Continuity of composite functions Prove Theorem 2.11: If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a),\) then the composition \(f \circ g\) is continuous at \(a\). (Hint: Write the definition of continuity for \(f\) and \(g\) separately; then combine them to form the definition of continuity for \(f \circ g .\) )

Step-by-Step Solution

Answer

Question: Prove Theorem 2.11: If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), then the composition \(f \circ g\) is continuous at \(a\). Answer: To prove Theorem 2.11, we combined the definitions of continuity for functions \(f\) and \(g\) and showed that the compositional function \(f \circ g\) satisfies the definition of continuity at point \(a\). We have shown that, for every \(\epsilon > 0\), there exists \(\delta_1 > 0\) such that \(|x - a| < \delta_1\) implies \(|f(g(x)) - f(g(a))| < \epsilon\). Thus, we have proven Theorem 2.11.

Step 1: Define continuity for functions f and g

We are given that \(f\) is continuous at \(g(a)\) and \(g\) is continuous at \(a\). By the definition of continuity for \(g\), for every \(\epsilon_1 > 0\), there exists \(\delta_1 > 0\) such that, for all \(x\) in the domain of \(g\): $$|x - a| < \delta_1 \Rightarrow |g(x) - g(a)| < \epsilon_1$$ Similarly, for continuity of \(f\) at \(g(a)\), for every \(\epsilon > 0\), there exists \(\delta > 0\) such that, for all \(y\) in the domain of \(f\): $$|y - g(a)| < \delta \Rightarrow |f(y) - f(g(a))| < \epsilon$$

Step 2: Combine definitions of continuity for f and g

Now we need to combine these definitions to find the continuity of \(f \circ g\) at \(a\). Consider the definition of continuity for \(f\), and let \(y = g(x)\): $$|g(x) - g(a)| < \delta \Rightarrow |f(g(x)) - f(g(a))| < \epsilon$$ Looking at the definition of continuity for \(g\), we can use an intermediate \(\epsilon_1 = \delta\): $$|x-a| < \delta_1 \Rightarrow |g(x) - g(a)| < \delta$$ Now, we can combine these definitions to form: $$|x - a| < \delta_1 \Rightarrow |f(g(x)) - f(g(a))| < \epsilon$$ This is the definition of continuity for the composition function \(f \circ g\) at point \(a\).

Step 3: Conclusion

We have shown that, for every \(\epsilon > 0\), there exists \(\delta_1 > 0\) such that \(|x - a| < \delta_1\) implies \(|f(g(x)) - f(g(a))| < \epsilon\), which satisfies the definition of continuity for the compositional function \(f \circ g\) at point \(a\). Thus, we have proven Theorem 2.11: If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), then the composition \(f \circ g\) is continuous at \(a\).

Key Concepts

Composite Functions

A composite function is formed by combining two functions such that the output of one function becomes the input of the other. Suppose you have two functions, \(f\) and \(g\). When you compose these functions, you create a new function denoted as \(f \circ g\). This means for any input \(x\), the expression \((f \circ g)(x) = f(g(x))\) describes the operation.

• The function \(g\) takes the input \(x\) and returns \(g(x)\).
• The function \(f\) then takes the result \(g(x)\) and returns \(f(g(x))\).

Understanding composite functions is essential because they appear frequently in calculus and various applications, such as physics and engineering. They allow us to model processes that happen in sequence, where the output of one stage is the input to the next.

Theorem 2.11

Theorem 2.11 discusses the continuity of composite functions. This theorem states that if you have two functions, \(g\) and \(f\), with \(g\) being continuous at some point \(a\) and \(f\) being continuous at \(g(a)\), then the composite function \(f \circ g\) is continuous at \(a\).The essence of this theorem lies in its step-by-step construction:

• First, recall that \(g\) is continuous at \(a\), meaning small changes in \(x\) around \(a\) produce small changes in \(g(x)\).
• Next, \(f\) being continuous at \(g(a)\) indicates that slight variations in \(g(x)\) around \(g(a)\) yield slight variations in \(f(g(x))\).
• By merging these two concepts, it follows that small changes around \(x = a\) result in small changes in the composite function \(f(g(x))\).

Theorem 2.11 is significant as it provides a powerful tool to assert the continuity of more complex functions built from simpler ones.

Definition of Continuity

The definition of continuity is a foundational concept in calculus. A function \(h\) is said to be continuous at a point \(c\) if for every real value \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(x\) is within \(\delta\) of \(c\), \(h(x)\) is within \(\epsilon\) of \(h(c)\).Mathematically, this is described as:\[\forall \epsilon > 0, \exists \delta > 0 \text{ such that if } |x - c| < \delta, \text{ then } |h(x) - h(c)| < \epsilon\]This definition ensures that as \(x\) approaches \(c\), \(h(x)\) approaches \(h(c)\) smoothly without any jumps or breaks. It’s a critical property for functions to possess, ensuring predictability and smooth transformations. Continuity has overarching implications, especially in understanding limits and integrals.

Epsilon-Delta Definition

The epsilon-delta (\(\epsilon-\delta\)) definition is a rigorous way to describe the concept of limits, and by extension, continuity. The framework provides precise conditions for limits that lead to defining when a function is continuous.

• "\(\epsilon\)" (\(\epsilon\)) represents a very small positive number that describes how close the function value \(f(x)\) should get to \(f(a)\) as \(x\) approaches \(a\).
• "\(\delta\)" (\(\delta\)) represents how close \(x\) should be to \(a\) to ensure \(f(x)\) is within \(\epsilon\) of \(f(a)\).

The core idea here is to show that no matter how tiny the range \(\epsilon\) is, you can find a corresponding \(\delta\) so that the function's value will stay within those bounds.This methodology not only defines continuity at a specific point but also allows mathematicians to analytically discuss and verify the behavior of functions in a structured way.