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The given problem asks to determine where the function is discontinuous for \( 0 \leq x \leq \frac{\pi}{2} \) and classify the discontinuities. The function provided is:

\[
f(x) = \frac{x^2 + 9x}{1 - \cos(3x)}
\]

### Step 1: Identify potential discontinuities
A function is discontinuous when the denominator is zero, as division by zero is undefined. For this function, the denominator is \( 1 - \cos(3x) \).

Set the denominator equal to zero:

\[
1 - \cos(3x) = 0
\]
\[
\cos(3x) = 1
\]

### Step 2: Solve for \( x \)
The cosine function equals 1 at integer multiples of \( 2\pi \). Hence:

\[
3x = 2n\pi \quad \text{(where \( n \) is an integer)}
\]
\[
x = \frac{2n\pi}{3}
\]

Now, restrict the values of \( x \) to the interval \( 0 \leq x \leq \frac{\pi}{2} \). For \( n = 0 \), we have:

\[
x = \frac{2(0)\pi}{3} = 0
\]

So, \( x = 0 \) is a point where the denominator equals zero.

### Step 3: Classify the discontinuity at \( x = 0 \)
At \( x = 0 \), check if the function is removable or non-removable.

The numerator at \( x = 0 \) is:

\[
x^2 + 9x = 0^2 + 9(0) = 0
\]

Both the numerator and denominator are zero at \( x = 0 \), indicating a potential removable discontinuity. To check, evaluate the limit of the function as \( x \) approaches 0.

\[
\lim_{x \to 0} \frac{x^2 + 9x}{1 - \cos(3x)}
\]

Using L'Hopital's Rule (since this is an indeterminate form \( \frac{0}{0} \)):

1. Differentiate the numerator:
   \[
   \frac{d}{dx} \left( x^2 + 9x \right) = 2x + 9
   \]
2. Differentiate the denominator:
   \[
   \frac{d}{dx} \left( 1 - \cos(3x) \right) = 3\sin(3x)
   \]

Now, compute the limit again:

\[
\lim_{x \to 0} \frac{2x + 9}{3\sin(3x)} = \frac{9}{0} = \infty
\]

Since the limit approaches infinity, the function has a non-removable infinite discontinuity at \( x = 0 \).

### Step 4: Conclusion
- The function is discontinuous at \( x = 0 \), and the discontinuity is an **infinite discontinuity**.
- There are no other discontinuities in the interval \( 0 < x \leq \frac{\pi}{2} \).

Would you like more details on any of these steps or further clarification?

### Related Questions:
1. How does L'Hopital's Rule help in determining the type of discontinuity?
2. What are the conditions for a function to have a removable discontinuity?
3. Can infinite discontinuities occur at points other than where the denominator is zero?
4. How would the analysis change if the interval of interest was larger than \( \frac{\pi}{2} \)?
5. Why does the cosine function affect the continuity of a rational function in this way?

**Tip:** When using L'Hopital's Rule, ensure the original expression is in the indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before differentiating.

Determine where the function is discontinuous for 0 ≤ x ≤ π/2 and classify the discontinuities. f(x) = (x^2 + 9x) / (1 - cos(3x))

Learn how to determine the discontinuities of the function f(x) = (x^2 + 9x) / (1 - cos(3x)) in the interval 0 ≤ x ≤ π/2. This solution involves trigonometric functions, limits, and L'Hopital's Rule.

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