Problem 9 - Calculus for Biology and Medicine

Question

In Problems 9-12, determine at which points \(f(x)\) is discontinuous. $$ f(x)=\frac{1}{x-3} $$

Step-by-Step Solution

Answer

The function is discontinuous at \( x = 3 \).

Step 1: Identify Potential Points of Discontinuity

The function \( f(x) = \frac{1}{x - 3} \) is a rational function, which means it can potentially be discontinuous at points where the denominator is zero. The denominator \( x - 3 \) equals zero when \( x = 3 \). This implies a potential point of discontinuity.

Step 2: Test for Discontinuity at Identified Points

To test whether \( f(x) \) is discontinuous at \( x = 3 \), consider the limit of \( f(x) \) as \( x \) approaches 3. As \( x \to 3 \), \( f(x) = \frac{1}{x-3} \) becomes undefined, because division by zero is not possible. Thus, the function is discontinuous at \( x = 3 \).

Key Concepts

Points of Discontinuity

When we discuss discontinuity in a function, we are referring to points where the function does not behave in a smooth or predictable manner. A common cause of discontinuity is where a function is undefined. Discontinuity points are particularly important when considering rational functions, which are composed of a numerator and a denominator. If the denominator equates to zero, this creates a division by zero scenario where the function cannot be evaluated.

In the function \( f(x) = \frac{1}{x-3} \), the denominator becomes zero at \( x = 3 \). This means \( f(x) \) is not defined at this point, clearly reflecting a point of discontinuity. Recognizing these points is crucial as they form gaps or breaks on the graph of the function, which can alter its overall behavior.

Rational Functions

Rational functions are expressions that can be written as the ratio of two polynomials. Mathematically, if \( f(x) = \frac{P(x)}{Q(x)} \), then \( f(x) \) is a rational function, where both \( P(x) \) and \( Q(x) \) are polynomials. An important aspect of rational functions is their potential points of discontinuity, caused by the zeros of the denominator.

For \( f(x) = \frac{1}{x-3} \), we see that it fits the form of a rational function where the numerator is \( 1 \) and the denominator is \( x - 3 \). The function's discontinuity at \( x = 3 \) is because substituting \( 3 \) into the denominator results in \( 0 \), making it undefined.

These discontinuities usually manifest as vertical asymptotes on the graph. When graphing, rational functions often appear smooth, except at these undefined points where the behavior can change suddenly and sharply.

Limit Evaluation

Limits are a fundamental concept in calculus used to analyze points of potential discontinuity. Specifically, they help us understand the behavior of a function as it approaches a certain point, even if the function itself is not defined there.

For example, with the function \( f(x) = \frac{1}{x-3} \), we are interested in the limit as \( x \to 3 \). Evaluating this limit shows how the function behaves near \( x = 3 \).

When \( x \) approaches 3 from either direction, the values of \( 1/(x-3) \) grow infinitely large in magnitude, as the denominator approaches zero. This confirms the function is discontinuous at this point, because no finite limit exists at \( x = 3 \). Such insights can determine the behavior of rational functions near their points of discontinuity and facilitate a deeper understanding of their graphs and applications.