Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

1. Limits and Continuity

Continuity

Calculus

1. Limits and Continuity

Continuity

Previous problemNext problem

1:51 minutes

Problem 2.5.13

Textbook Question

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

Verified step by step guidance

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 \( -3x \).

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: \( x - 2 = 0 \).

Solve the equation \( x - 2 = 0 \) to find the point of discontinuity. This gives \( x = 2 \). Therefore, the function is discontinuous at \( x = 2 \).

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.

Conclude that the function \( y = \frac{1}{x - 2} - 3x \) is continuous for all real numbers except \( x = 2 \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.