The question in the image asks:

Q1. At what points are functions continuous? Given function: $y = \frac{1}{x-2} - 3x$

To determine the continuity of the function:

1. The function is the sum of two parts: $\frac{1}{x-2}$ and $-3x$.

   • The term $-3x$ is a polynomial, and polynomials are continuous everywhere in the real number system.
   • The term $\frac{1}{x-2}$ is a rational function, which is continuous everywhere except where the denominator is zero.

2. The denominator of the rational term $\frac{1}{x-2}$ is zero when $x = 2$. At this point, the function is undefined, so the function is discontinuous at $x = 2$.

Conclusion:

• The function $y = \frac{1}{x-2} - 3x$ is continuous for all real numbers except at $x = 2$, where it has a discontinuity.

Would you like more details on the nature of this discontinuity or further clarification?

Related questions:

1. What is the definition of a continuous function at a point?
2. How do you identify discontinuities in a rational function?
3. What types of discontinuities exist in functions?
4. Can a function be made continuous at a discontinuity? If so, how?
5. How does the behavior of a function near a discontinuity affect its limit?

Tip:

When examining rational functions, always check where the denominator equals zero, as this typically indicates points of discontinuity.