Problem 26 - Contemporary Precalculus

Question

Determine all numbers at which the function is continuous. $$g(x)=\left\\{\begin{array}{ll} \frac{x^{2}-x-6}{x^{2}-4} & \text { if } x \neq-2 \\ 5 / 4 & \text { if } x=-2 \end{array}\right.$$

Step-by-Step Solution

Answer

The function g(x) is continuous for all numbers, except for x = 2, as the denominator becomes zero at x = 2.

Step 1: Check continuity for x ≠ -2

As the function is a rational function for all x ≠ -2, we only need to check if the denominator is non-zero to determine continuity at these points. The denominator of the function is given by: \(x^{2} - 4\) This denominator is non-zero for all x ≠ -2, except for x = 2 (since at x = 2, the denominator becomes zero). Therefore, the function is continuous for all x ≠ -2, except at x = 2.

Step 2: Check continuity for x = -2

At x = -2, the function is defined as a constant value 5/4. To check continuity at x = -2, we need to evaluate the limits from the left of -2 and from the right of -2: - Limit from the left: \(\displaystyle\lim_{x\to -2^{-}} g(x)\) - Limit from the right: \(\displaystyle\lim_{x\to -2^{+}} g(x)\) If these two limits exist and are equal, the function is continuous at x = -2.

Step 3: Evaluate the limits

Let's first find the limit from the left. For x < -2, the function is given by: \(g(x) = \frac{x^{2}-x-6}{x^{2}-4}\) So, the limit from the left is: \(\displaystyle\lim_{x\to -2^{-}} g(x) =\lim_{x\to -2^{-}} \frac{x^{2}-x-6}{x^{2}-4}\) Similarly, let's find the limit from the right. For x > -2, the function is still given by: \(g(x) = \frac{x^{2}-x-6}{x^{2}-4}\) So, the limit from the right is: \(\displaystyle\lim_{x\to -2^{+}} g(x) =\lim_{x\to -2^{+}} \frac{x^{2}-x-6}{x^{2}-4}\) Now, since the function is the same for x < -2 and x > -2, the limits from the left and the right will be equal. Let's evaluate the limit: \(\displaystyle\lim_{x\to -2} \frac{x^{2}-x-6}{x^{2}-4} = \frac{(-2)^{2}-(-2)-6}{(-2)^{2}-4} = \frac{4+2-6}{4-4} = \frac{0}{0}\) The limit is in an indeterminate form. We can apply factoring to simplify the limit.

Step 4: Simplify the limit

We can factor the numerator and the denominator of the limit to simplify it: Numerator: \(x^{2} - x -6 = (x+2)(x-3)\) Denominator: \(x^{2} - 4 = (x-2)(x+2)\) So, the limit becomes: \(\displaystyle\lim_{x\to -2} \frac{(x+2)(x-3)}{(x-2)(x+2)}\) Now, we can cancel out the (x+2) term from the numerator and the denominator: \(\displaystyle\lim_{x\to -2} \frac{x-3}{x-2}\)

Step 5: Evaluate the simplified limit

Now, we can evaluate the limit with the simplified expression: \(\displaystyle\lim_{x\to -2} \frac{x-3}{x-2} = \frac{-2-3}{-2-2} = \frac{5}{4}\) Since the limit for x approaching -2 exists and is equal to 5/4, and the function is also defined to be 5/4 at x=-2, the function is continuous at x = -2.

Step 6: Conclusion

The function g(x) is continuous for all x ≠ -2, except at x = 2 (as the denominator is zero). Moreover, g(x) is continuous at x = -2, as the limit from both sides exists and equals the value of the function at x = -2. Therefore, the function is continuous for all numbers, except for x = 2.

Key Concepts

Rational Functions

Rational functions are mathematical expressions that represent the ratio of two polynomials. A typical rational function is represented as \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials and \( q(x) \) is not equal to zero. The points where the denominator \( q(x) = 0 \) are of special interest because they are often where the function is undefined and can lead to discontinuities.

In the context of continuity, if a rational function has no points where the denominator equals zero, it is considered continuous over its entire domain. However, real-world problems, like the exercise provided, involve functions that may have a constant value at some points, altering the analysis for continuity.

Limits

Limits are fundamental to the study of calculus and analysis. They describe the behavior of a function as the input approaches a certain value. For a function \( f(x) \), the limit as \( x \) approaches a value \( c \) is represented as \( \lim_{x\to c} f(x) \). If the limit exists and equals the function's value at that point, the function is continuous at that point.

In our exercise, the limits from the left and the right of \( x = -2 \) determine the function's continuity at that point. If these limits are equal to each other and to the function's value at \( x = -2 \), the function is continuous there. This concept is crucial for understanding when and why a function might exhibit discontinuities.

Indeterminate Forms

Indeterminate forms occur when evaluating a limit leads to an expression like \( \frac{0}{0} \) or \( \infty / \infty \), which does not produce a definitive value. These forms do not mean the limit does not exist; rather, they signal that further steps are necessary to find the limit's value.

Factoring, L'Hôpital's rule, and other algebraic techniques can often resolve indeterminate forms. In the exercise, once we identify the indeterminate form \( \frac{0}{0} \), we factor the numerator and denominator to cancel common terms and evaluate the simplified limit. This step is crucial to determining the function's behavior and its continuity at the point \( x = -2 \).