At what points are the functions in Exercises 13–30 continuous?
y = 1/(x – 2) – 3x

Verified step by step guidance

1. Identify the function given: \( y = \frac{1}{x - 2} - 3x \). This function is composed of two parts: a rational function \( \frac{1}{x - 2} \) and a linear function \( -3x \).
2. Determine the points of discontinuity for the rational function \( \frac{1}{x - 2} \). A rational function is discontinuous where its denominator is zero. Set the denominator equal to zero: \( x - 2 = 0 \).
3. Solve the equation \( x - 2 = 0 \) to find the point of discontinuity. This gives \( x = 2 \). Therefore, the function is discontinuous at \( x = 2 \).
4. Consider the linear function \( -3x \). Linear functions are continuous everywhere on the real number line, so there are no additional points of discontinuity from this part of the function.
5. Conclude that the function \( y = \frac{1}{x - 2} - 3x \) is continuous for all real numbers except \( x = 2 \).

Key Concepts

Continuity of Functions

A function is continuous at a point if it is defined at that point, the limit exists at that point, and the limit equals the function's value at that point. For rational functions, continuity is typically disrupted by points where the denominator equals zero, leading to undefined values.

Identifying Discontinuities

To determine where a function is continuous, one must identify points of discontinuity, which occur when the denominator of a rational function is zero. For the given function, setting the denominator (x - 2) to zero reveals potential discontinuities, which must be analyzed further.

Limits

Limits are fundamental in calculus for understanding the behavior of functions as they approach specific points. Evaluating limits helps determine if a function approaches a finite value or diverges, which is crucial for assessing continuity at points where the function may be undefined.