Problem 13 - Calculus Early Transcendentals

Question

Continuity at a point Determine whether the following functions are continuous at a. Use the continuity checklist to justify your answer. $$f(x)=\frac{2 x^{2}+3 x+1}{x^{2}+5 x} ; a=5$$

Step-by-Step Solution

Answer

Answer: Yes, the function is continuous at the point a=5.

Step 1: Check if the function is defined at the point a=5

In order to check if the function is defined at the point a=5, we need to substitute a=5 into the function and check if it is possible to compute the function value: $$f(5)=\frac{2(5)^{2}+3(5)+1}{(5)^{2}+5(5)}$$ Since the denominator is not equal to zero, the function is defined at the point a=5.

Step 2: Check if the limit exists as x approaches a=5

To check if the limit of the function exists as x approaches a=5, we have to check the limit from both the left side and the right side. $$\lim_{x\to 5} f(x) = \lim_{x\to 5} \frac{2x^2+3x+1}{x^2+5x}$$ By checking the rational function, we observe that the degree of the numerator and the degree of the denominator are equal. In this case, the limit exists and can be calculated by dividing the leading coefficients. $$\lim_{x\to 5} f(x) = \frac{2}{1} = 2$$

Step 3: Check if the limit equals the function value at the point a=5

We have already calculated the function value at a=5 in Step 1, and the limit as x approaches a=5 in Step 2. Now, we need to compare them to check if they are equal. _function value at a=5:_ $$f(5) =\frac{2(5)^{2}+3(5)+1}{(5)^{2}+5(5)}$$ _Limit as x approaches a=5:_ $$\lim_{x\to 5} f(x) = 2$$ Calculate the function value at a=5: $$f(5) = \frac{2(25)+3(5)+1}{(25)+5(5)}=\frac{50+15+1}{25+25}=\frac{66}{50}=2$$ Since the function value at a=5 and the limit as x approaches a=5 are equal, we can conclude that the function is continuous at the point a=5.

Key Concepts

Limit of a function

When we talk about the "limit of a function," we are trying to understand what value the output of a function approaches as the input gets closer to a specific point. It's like asking: "What happens to \( f(x) \) when \( x \) gets really close to 5, but not exactly 5?" This approach helps us analyze the behavior of functions near a point, which is crucial for determining continuity.
For the function \( f(x) = \frac{2x^2+3x+1}{x^2+5x} \), we analyzed what happens as \( x \) gets close to 5. We found that:

• The limit from the left (as \( x \) approaches 5 from values less than 5) equals the limit from the right (as \( x \) approaches 5 from values greater than 5).
• Both of these limits tell us the function approaches a common value, which means the general limit exists at \( x = 5 \).

In mathematical terms: \[\lim_{x \to 5} \frac{2x^2+3x+1}{x^2+5x} = 2\]Finding the limit is the beginning of determining if a function is continuous at a point.

Function evaluation

Function evaluation is the process of substituting a specific value into a function to find its output. This step is key when checking continuity at a point because it confirms that the function output is well-defined.
In our example, the next task was to evaluate the function at \( x = 5 \). We did this by directly substituting 5 into the function \( f(x) = \frac{2x^2+3x+1}{x^2+5x} \). The result was simple:

• Calculate the numerator: you take \( 2(5)^2 + 3(5) + 1 \).
• Calculate the denominator: you take \( (5)^2 + 5(5) \).
• Then, divide the results: \( \frac{66}{50} \) simplifies to \( 2 \).

This shows us that \( f(5) = 2 \). Crucially, it confirms two things: first, that the function is defined at this point, and second, it gives a concrete value to compare against the computed limit to check for continuity.

Rational functions

Rational functions are a type of function expressed as the ratio of two polynomials. They often appear as fractions with a polynomial in both the numerator and the denominator. Understanding them is pivotal in calculus, particularly when evaluating limits and continuity.
In our specific function \( f(x) = \frac{2x^2+3x+1}{x^2+5x} \), both the numerator and the denominator are polynomial expressions. Here are some key characteristics:

• They can have vertical asymptotes where the denominator equals zero.
• They can be analyzed for end behavior using limits.
• They are well-suited to analysis at different points through substitution and limit evaluation.

For our exercise, since the denominator never becomes zero at \( x = 5 \), it behaves nicely—allowing direct substitution for function evaluation. However, we see how checking the limits helps predict behavior, especially because rational functions can misbehave at critical points (where the denominator might approach zero). Thus, rational functions require specific attention when determining limits and checking continuity.