Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

State whether the indicated function is continuous at \(3 .\) If it is not continuous, tell why. $$ h(x)=\frac{x^{2}-9}{x-3} $$

Step-by-Step Solution

Answer

The function is not continuous at \(x = 3\) because it is not defined at that point.

Step 1: Identify Points of Discontinuity

To determine the points of discontinuity, we need to check if the function is defined at the point of interest, which is at \(x = 3\). Substitute \(x = 3\) into the denominator to see if it causes division by zero.\[ \text{Denominator at } x = 3: x - 3 = 3 - 3 = 0 \]Since the denominator is zero, the function is not defined at \(x = 3\), indicating a possible discontinuity.

Step 2: Factor the Numerator

To understand why there is a discontinuity at \(x = 3\), factor the numerator of the function.\[ h(x) = \frac{x^{2}-9}{x-3} \]The numerator \(x^2 - 9\) is a difference of squares:\[ x^2 - 9 = (x - 3)(x + 3) \]So the function becomes:\[ h(x) = \frac{(x-3)(x+3)}{x-3} \]

Step 3: Simplify the Function

Simplify the function by canceling the common term \((x - 3)\) present in both the numerator and the denominator:\[ h(x) = x + 3 \]However, \(x = 3\) was an original restriction due to division by zero, so the simplified function \(h(x) = x + 3\) is valid only for \(x eq 3\).

Step 4: Evaluate the Function's Limit at x = 3

Even though the function is not defined at \(x = 3\), its limit as \(x\) approaches 3 can be evaluated using the simplified form:\[ \lim_{{x \to 3}} (x + 3) = 3 + 3 = 6 \]

Step 5: Determine Continuity

A function is continuous at a point if it is defined at that point, and if its limit as \(x\) approaches the point equals the function's value at that point. Here, \(h(x)\) is not defined at \(x = 3\), therefore \(h(x)\) is not continuous at \(x = 3\).

Key Concepts

Points of Discontinuity

In calculus, identifying points of discontinuity is crucial for understanding the behavior of functions. A function is said to be discontinuous at a point if it is not defined, or if it behaves erratically near that point. In the original exercise, we looked at the function \(h(x) = \frac{x^2 - 9}{x-3}\) to determine if it's continuous at \(x = 3\). Discontinuity often occurs when the denominator of a rational function equals zero, leading to undefined values.
In this particular case, plugging \(x = 3\) into the denominator results in zero: \(x - 3 = 0\). As the function becomes undefined, this signals a discontinuity at \(x = 3\). Knowing where these points of discontinuity lie helps in analyzing and graphing functions correctly.

Limit Evaluation

Evaluating limits helps us understand the behavior of functions as they approach certain points, even if they are not defined at those points. For our example, we used the simplified form \(h(x) = x + 3\) after canceling common factors except at \(x = 3\).
The limit of \(h(x) = x + 3\) as \(x\) approaches 3 is evaluated as follows:

• Substitute the approaching value into the simplified expression: \(\lim_{{x \to 3}} (x + 3)\).
• This leads to \(3 + 3 = 6\).

This evaluation gives us the limit value 6, showing how the function behaves near the point of discontinuity.

Factorization

Factorization is breaking down expressions into simpler components that can be multiplied to obtain the original expression. In our given function, factorization was used to simplify \(x^2 - 9\).
This polynomial expression can be factored as a difference of squares, given by:

• \(x^2 - 9 = (x - 3)(x + 3)\).

Factorization helps in simplifying expressions and identifying removable discontinuities, effectively "canceling out" terms in rational functions like in our exercise. This step is vital for simplifying the function and understanding its fundamental structure.

Difference of Squares

The concept of difference of squares is a special algebraic identity used in factorization. Recognizing this pattern makes simplifying expressions much easier. It takes the form:

• \(a^2 - b^2 = (a - b)(a + b)\).

In our function \(h(x) = \frac{x^2 - 9}{x-3}\), the numerator \(x^2 - 9\) fits the difference of squares pattern because it can be expressed as \((x - 3)(x + 3)\). By applying this identity correctly, we facilitated the simplification of the function.
Understanding and applying the difference of squares allows for more efficient problem solving in algebra and calculus, particularly when dealing with polynomial expressions.