Problem 13 - Calculus Early Transcendentals

Question

\(13-14\) Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval. \(f(x)=\frac{2 x+3}{x-2},(2, \infty)\)

Step-by-Step Solution

Answer

The function is continuous on \((2, \infty)\).

Step 1: Define Continuity at a Point

A function \( f(x) \) is continuous at a point \( c \) if the following conditions are met:\1. \( f(c) \) is defined.\2. The limit \( \lim_{x \to c} f(x) \) exists.\3. \( \lim_{x \to c} f(x) = f(c) \).\For \( f(x) = \frac{2x + 3}{x - 2} \), we want to show continuity for any point \( c \) in the interval \( (2, \infty) \).

Step 2: Check Function Definition on Interval

The function \( f(x) = \frac{2x + 3}{x - 2} \) is defined for all values of \( x \) except when \( x = 2 \). Since the interval is \( (2, \infty) \), \( f(c) \) is defined for each \( c \) in the interval.

Step 3: Evaluate the Limit as x Approaches c

To ensure continuity at a point \( c \) in \( (2, \infty) \), evaluate the limit \( \lim_{x \to c} \frac{2x + 3}{x - 2} \) using limit properties. Since the function is a rational function and \( x eq 2 \) for our interval, apply direct substitution: \[ \lim_{x \to c} \frac{2x + 3}{x - 2} = \frac{2c + 3}{c - 2}. \]

Step 4: Verify the Limit Equals the Function Value

Since \( f(c) = \frac{2c + 3}{c - 2} \), the limit \( \lim_{x \to c} f(x) = \frac{2c + 3}{c - 2} = f(c) \). Thus, the limit equals the function value at each \( c \) in \( (2, \infty) \).

Step 5: Conclude Continuity

All conditions for continuity are satisfied for every \( c \) in the interval \( (2, \infty) \):\- \( f(c) \) is defined,\- \( \lim_{x \to c} f(x) \) exists,\- \( \lim_{x \to c} f(x) = f(c) \). The function \( f(x) = \frac{2x + 3}{x - 2} \) is continuous on \( (2, \infty) \).

Key Concepts

Limits

Limits are a fundamental concept in calculus that help us understand the behavior of functions as they approach certain points. Essentially, a limit examines what happens to a function as the input value approaches a specific point. In mathematical notation, we express a limit as \( \lim_{x \to c} f(x) \), which reads as "the limit of \( f(x) \) as \( x \) approaches \( c \)."

When we deal with rational functions, which are quotients of polynomials, limits help determine how the function behaves near values that make the denominator zero. For the function \( f(x) = \frac{2x + 3}{x - 2} \), calculating the limit as \( x \) approaches any point in the interval \((2, \infty)\) shows us how the function behaves as it gets very close to that point without actually being exactly at the undefined value \( x=2 \).

• If a function is approaching a specific value as \( x \to c \) and does so consistently, we can say the limit exists.
• Having a definable limit is crucial in discussing continuity of a function.

Rational Functions

Rational functions, as mentioned earlier, are quotients of two polynomials. These functions are written in the form \( f(x) = \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials.

In our example, \( f(x) = \frac{2x + 3}{x - 2} \), the function is undefined at \( x = 2 \) due to division by zero, which means the function has a vertical asymptote at that point. As long as \( x \) remains distinct from 2, the function is well-behaved, allowing us to explore its attributes like continuity over other intervals.

It's essential to remember:

• Rational functions are continuous across their domains unless affected by divisions by zero or other constraints.
• We investigate these restrictions to understand how they transform the graph of the function around troublesome points.

Continuous Functions

A function is continuous at a point if it doesn't have any abrupt changes in value around that point. For a function to be continuous at a point \( c \), three conditions must hold:
1. \( f(c) \) should be defined, meaning \( c \) should be within the domain of the function.
2. The limit \( \lim_{x \to c} f(x) \) should exist.
3. The limit must equal the function value at that point, i.e., \( \lim_{x \to c} f(x) = f(c) \).

For our given function, continuity was demonstrated on the interval \((2, \infty)\) by:

• Confirming that \( f(x) = \frac{2x + 3}{x - 2} \) is defined for any \( c \) in this interval.
• Ensuring the limit \( \lim_{x \to c} \frac{2x + 3}{x - 2} \) exists and matches the function value \( f(c) \).

These steps establish that the function flows smoothly within this interval without any breaks or jumps.

Calculus

Calculus is a branch of mathematics that allows us to explore changes, rates, and behaviors of different mathematical structures. It is essential for analyzing and solving complex real-world problems. Two of the foundational concepts in calculus are limits and continuity.

The full power of calculus is seen when these concepts are used to study not only the behavior of functions at specific points but also how these functions change and provide insight into more extensive problems. Concepts like differential calculus, which deals with how a function changes (derivatives), and integral calculus, which concerns areas under curves and accumulation (integrals), build off continuity and limits.

• Continuity ensures a function behaves predictably, which is crucial for further analyses.
• Limits help define these changes mathematically, allowing deeper investigation through calculus techniques.

In this specific exercise, understanding these basics allows us to confirm and articulate why \( f(x) = \frac{2x + 3}{x - 2} \) behaves continually and predictably across the interval \((2, \infty)\).