The image contains multiple mathematical functions, and the task is to determine at which points each function is continuous. I will analyze each function and explain where they may fail to be continuous.

1. i) $y = \frac{1}{x - 2} - 3x$

   • This is a rational function. It is continuous everywhere except where the denominator is zero.
   • Discontinuity: $x = 2$.

2. ii) $y = \frac{1}{(x + 2)^2} + 4$

   • This is also a rational function. It is continuous except where the denominator is zero.
   • Discontinuity: $x = -2$ (since $(x + 2)^2 = 0$ at $x = -2$).

3. iii) $y = \frac{x + 1}{x^2 - 4x + 3}$

   • The denominator factors as $(x - 3)(x - 1)$, so the function is continuous except where the denominator is zero.
   • Discontinuities: $x = 1$ and $x = 3$.

4. iv) $y = \frac{x + 3}{x^2 - 3x + 10}$

   • The denominator $x^2 - 3x + 10$ has no real roots because the discriminant $(-3)^2 - 4(1)(10) = -31$ is negative.
   • No discontinuities, continuous for all real $x$.

5. v) $y = |x - 1| + \sin(x)$

   • The absolute value function and sine function are continuous everywhere.
   • Continuous everywhere.

6. vi) $y = \frac{1}{|x| + 1} - \frac{x^2}{2}$

   • The first term $\frac{1}{|x| + 1}$ is continuous because $|x| + 1 \neq 0$ for all real $x$. The second term $\frac{x^2}{2}$ is a polynomial, so it's also continuous everywhere.
   • Continuous everywhere.

7. vii) $y = \frac{\cos(x)}{x}$

   • This is a rational function. It is continuous everywhere except where the denominator is zero.
   • Discontinuity: $x = 0$.

8. viii) $y = \frac{x + 2}{\cos(x)}$

   • This is also a rational function. It is continuous except where the denominator $\cos(x) = 0$, which happens at $x = \frac{\pi}{2} + n\pi$ for integers $n$.
   • Discontinuities: $x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$.

9. ix) $g(x) = \begin{cases} x^2 - x - 6 & x \neq 3 \\ 5 & x = 3 \end{cases}$

   • The function is piecewise. For $x \neq 3$, the function is $x^2 - x - 6 = (x - 3)(x + 2)$, which is continuous for all $x$. However, at $x = 3$, the function takes the value 5, which is different from the value of $(x - 3)(x + 2)$ at $x = 3$, which would be $0$.
   • Discontinuity: $x = 3$.

Summary of Discontinuities:

• i) $x = 2$
• ii) $x = -2$
• iii) $x = 1, x = 3$
• iv) None
• v) None
• vi) None
• vii) $x = 0$
• viii) $x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$
• ix) $x = 3$

Would you like more details on any of these functions or their continuity? Here are five related questions to further explore:

1. What does the Intermediate Value Theorem say about continuous functions?
2. How do we determine the type of discontinuity (removable, jump, etc.)?
3. How does continuity affect the differentiability of a function?
4. Why are rational functions often discontinuous at specific points?
5. Can we make a discontinuous function continuous by redefining a point?

Tip: The absolute value function $|x|$ is continuous everywhere, but its derivative has a sharp corner at $x = 0$, making it non-differentiable at that point.