Problem 23 - Calculus Early Transcendentals

Question

Use Theorem 2.10 to determine the intervals on which the following functions are continuous. $$f(x)=\frac{x^{5}+6 x+17}{x^{2}-9}$$

Step-by-Step Solution

Answer

Answer: $$(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$$

Step 1: Identify the denominator

In this function, $$f(x) = \frac{x^{5}+6 x+17}{x^{2}-9}$$, the denominator is the expression $$x^{2} - 9$$.

Step 2: Determine when the denominator is equal to zero

To determine the values of x that make the denominator equal to zero, set $$x^{2} - 9 = 0$$. $$x^{2}-9=0$$ $$(x+3)(x-3)=0$$ $$x=-3, x=3$$ So, the denominator is equal to zero when $$x = -3$$ and $$x = 3$$.

Step 3: Determine the intervals of continuity

Since a rational function is continuous everywhere except at the values that make the denominator equal to zero, the function is continuous on the intervals where x is less than -3, between -3 and 3, and greater than 3. This can be expressed as: $$(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$$

Step 4: Conclusion

The function $$f(x) = \frac{x^{5}+6 x+17}{x^{2}-9}$$ is continuous on the intervals $$(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$$.

Key Concepts

Continuous Functions

Understanding the concept of continuous functions is crucial for analyzing how a function behaves within certain intervals. Imagine you draw a curve on a graph without lifting your pencil – that's essentially what it means for a function to be continuous over an interval. In a more technical sense, a continuous function is one where small changes in the input result in small changes in the output. The graph of such a function forms an unbroken curve.

For a function to be continuous at a point, it must satisfy three criteria: the function must be defined at the point, the limit as the function approaches that point must exist, and the function's value at that point must be equal to that limit. When dealing with polynomial functions, like the numerator of our exercise function, they are always continuous. However, complexity arises with functions that have defined denominators, such as rational functions, where continuity might be interrupted.

Rational Functions

Rational functions are quite interesting in the way they mirror the ratio of two polynomials – much like a fraction. As seen in the exercise function \( f(x) = \frac{x^{5}+6 x+17}{x^{2}-9} \), the numerator is a polynomial, and so is the denominator. The key characteristic that affects the continuity of a rational function is the points where the denominator equals zero – these are points of potential discontinuity. In our exercise, \( x^{2} - 9 = 0 \) when \( x = -3 \) or \( x = 3 \), creating discontinuities there.

Anywhere else, the function behaves nicely and continues as usual. That's why we define the intervals of continuity for \( f(x) \) without the values that cause the denominator to vanish into the mathematical void of division by zero. It’s important to identify the non-permissible values to understand the full scope of the function's behaviour.

Theorem Application

In mathematics, applying theorems correctly allows us to deduce properties about functions systematically. The exercise in question refers to Theorem 2.10, which likely informs us about the continuity of rational functions. While the specifics of the theorem aren't detailed in the problem, one can infer that it states a rational function is continuous wherever its denominator is non-zero.

By using this theorem, identifying the zeros of the denominator function, and excluding them, we determined the intervals of continuity for the rational function \( f(x) \). Being able to apply such theorems not only guides us to the correct answer quickly but also reinforces a deeper understanding of the function's nature and when it can be used without encountering infinity or undefined values.