Let's discuss the continuity of the given function:

$f(x) = \frac{x + 2}{x - 3}$

1. Continuity on the open interval $(-2 < x < 3)$

• The function $f(x)$ is a rational function, which is continuous everywhere in its domain (where the denominator is not zero).
• The function has a discontinuity where the denominator $x - 3 = 0$, i.e., at $x = 3$, because dividing by zero is undefined.
• On the interval $(-2, 3)$, the denominator does not become zero, so the function is continuous everywhere on this open interval.

2. Continuity on the closed interval $[-2 \leq x \leq 3]$

• The closed interval includes the point $x = 3$, which is a point of discontinuity for the function.
• Therefore, $f(x)$ is not continuous on the closed interval $[-2, 3]$ because of the discontinuity at $x = 3$.

Conclusion:

• On the open interval $(-2 < x < 3)$: The function is continuous.
• On the closed interval $[-2 \leq x \leq 3]$: The function is not continuous because it has a discontinuity at $x = 3$.

Would you like further details on this or have any other questions?

Here are 5 related questions for deeper exploration:

1. What is a removable discontinuity, and does $x = 3$ represent one?
2. How can we find the limit of $f(x)$ as $x$ approaches 3?
3. What happens to the value of $f(x)$ as $x$ approaches 3 from the left and right?
4. What does it mean for a function to be continuous at a point?
5. How would you graph this function, showing its points of discontinuity?

Tip: When checking for continuity, always evaluate the domain of the function first, and then identify where it might fail to be continuous.