Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

State whether the indicated function is continuous at \(3 .\) If it is not continuous, tell why. $$ f(x)=\frac{21-7 x}{x-3} $$

Step-by-Step Solution

Answer

The function is not continuous at \(x = 3\) due to division by zero.

Step 1: Check for continuity at the point

To determine if the function is continuous at a given point, first check whether the function is defined at that point. The function is continuous at a point if it is defined at that point, the limit of the function exists there, and the limit equals the function's value.

Step 2: Determine if the function is defined at the point

Evaluate the function at the point of interest, which is 3 in this case:\[f(3) = \frac{21 - 7 \times 3}{3 - 3} = \frac{0}{0}\]The function is not defined at \(x = 3\) because it results in a division by zero, indicating a discontinuity.

Step 3: Find the limit as x approaches 3

Simplify and find the limit of \(f(x)\) as \(x\) approaches 3:\[\lim_{{x \to 3}} \frac{21-7x}{x-3}\]Factor the numerator:\[\lim_{{x \to 3}} \frac{7(3-x)}{x-3} = \lim_{{x \to 3}} \frac{-7(x-3)}{x-3}\]Now, cancel \(x-3\) terms:\[\lim_{{x \to 3}} -7 = -7\]

Step 4: Determine continuity

Since the function is not defined at \(x = 3\), despite the limit existing as \(x\) approaches 3, the function \(f(x)\) is not continuous at \(x = 3\).

Key Concepts

Limits and Continuity

Understanding limits and continuity is a fundamental aspect of calculus. A function is said to be continuous at a point if the value of the function at that point is equal to the limit of the function as it approaches the point from both directions.

For a function to be continuous at a point, three conditions must be satisfied:

• The function is defined at the point.
• The limit of the function exists as the variable approaches the point.
• The value of the function at that point equals the limit.

In the given exercise, the function is not continuous at the point 3 because it is not defined at that point, which results in a crucial failure to meet one of these conditions.

Discontinuity

Discontinuity occurs when a function is not continuous at a particular point. This can happen for various reasons, including division by zero or an undefined expression.

In the exercise, the function \( f(x) = \frac{21-7x}{x-3} \) shows a discontinuity at \( x = 3 \) because substituting \( x = 3 \) leads to division by zero, yielding an undefined expression.

Discontinuities can be classified into different types, such as removable, jump, or infinite discontinuities. In this case, we have a removable discontinuity because the function can be simplified to eliminate the factors that cause the undefined point. By canceling out the terms, the limit at the point exists, indicating the type of discontinuity.

Rational Functions

Rational functions are functions that can be expressed as the quotient of two polynomials. They often take the form \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials.

One of the key characteristics of rational functions is that they might have points where they are undefined, particularly where the denominator equals zero.

When dealing with rational functions, it's important to identify the values that make the denominator zero, as these are potential points of discontinuity.

For the function in the exercise, \( f(x) = \frac{21-7x}{x-3} \), the denominator becomes zero when \( x = 3 \), leading to a division by zero and an undefined function at that point. Simplifying rational functions by factoring is a common method to deal with such issues and potentially find limits at these problematic points.