Problem 17 - 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

Explain your answer. Answer: No, the function is not continuous at x = -5 because f(a) is not defined at this point, as the denominator becomes 0.

Step 1: Check if f(a) is defined

Firstly, we need to check if f(-5) exists. To do this, we plug in x = -5 into the given function: $$f(-5) = \frac{2(-5)^2 + 3(-5) + 1}{(-5)^2 + 5(-5)}$$ $$f(-5) = \frac{50 - 15 + 1}{25 - 25}$$ As we can see, the denominator is 0, which means we cannot find a value for f(-5). Therefore, f(a) is not defined, and the function is not continuous at x = -5.

Step 2: Analyzing the lack of continuity

Since f(a) is not defined, we don't even need to check the other two criteria in the continuity checklist. We can already conclude that the function is not continuous at x = -5. However, for further understanding, let's analyze the problematic term that led to the denominator being 0: $$x^{2} + 5x = x(x + 5)$$ Notice that this term becomes 0 when x = 0 or x = -5. In other words, the function will have a discontinuity at these points.

Step 3: Final conclusion

The given function $$f(x)=\frac{2 x^{2}+3 x+1}{x^{2}+5 x}$$ is not continuous at x = -5, as we found that f(a) is not defined. Therefore, the function does not meet the criteria for continuity at this point.

Key Concepts

Continuous Function

A continuous function is a mathematical function that has no breaks, gaps, or jumps when you examine its graph. Imagine drawing the curve of the function on a piece of paper without lifting your pencil; if you can do this for the entire range of the function, then it is continuous.

In more technical terms, a function is continuous at a point 'a' if three key conditions are met:

1. The function is defined at 'a' (\( f(a) \text{ exists} \)).
2. The limit of the function as it approaches 'a' exists (\( \text{lim}_{x \to a} f(x) \text{ exists} \)).
3. The function's value at 'a' and the limit as x approaches 'a' are the same (\( f(a) = \text{lim}_{x \to a} f(x) \)).

When a function does not meet all these conditions, it's not continuous at that specific point. In the given exercise, the function fails to be continuous at the point 'a = -5' because the very first condition is not satisfied: the function is not defined at 'a' because it leads to division by zero, which is undefined in mathematics.

Discontinuity

The concept of discontinuity in calculus refers to points where a function fails to be continuous. When a function is not continuous at a particular point, it means there's some kind of interruption in the behavior of the function at that point.

There are several types of discontinuities:

• Point discontinuity occurs when a function is undefined or defined differently at a single point.
• Jump discontinuity happens when there is a sudden 'jump' in the function values at a point.
• Infinite discontinuity is when values become infinitely large near a certain point.
• Oscillatory discontinuity occurs when function values oscillate wildly near a point.
• Removable discontinuity is a type that can be 'fixed' by redefining the function at a point.

In our exercise, we deal with a type of discontinuity at 'x = -5' arising because the value of the function at that point is not defined – this can be considered as a point discontinuity.

Limits in Calculus

The concept of limits is foundational in calculus and entails understanding the behavior of a function as the input value approaches a particular point. A limit can predict a function's value at a specific point, even if the function itself is not actually defined at that point.

Mathematically, the notation

\( \text{lim}_{x \to a} f(x) = L \)

signifies that as 'x' gets very close to 'a', the function 'f(x)' approaches the value ‘L’. Limits are crucial in determining the continuity of a function at a point, as they help to specify the function's behavior around that point.

In our example, due to the function not being defined at 'x = -5', we cannot calculate its limit at that point. Not having this limit further cements the fact that we have a discontinuity at 'x = -5'. Limits are also used to calculate derivatives and integrals, which are central to the study of calculus.