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We are given a function \( f(x) \) that is piecewise defined on the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), and we are asked to determine the constants \( a \) and \( b \) such that \( f(x) \) is continuous at \( x = 0 \).

### The function is given as:

\[
f(x) =
\begin{cases}
\frac{3x^2}{\sqrt{x^2 + a^2} - a}, & \text{if } -\frac{\pi}{2} < x < 0 \\
b, & \text{if } x = 0 \\
\frac{b}{3^{1 - \tan x} + 3^{1 - \cot x}}, & \text{if } 0 < x < \frac{\pi}{2}
\end{cases}
\]

To ensure the continuity of \( f(x) \) at \( x = 0 \), the left-hand limit \( \lim_{x \to 0^-} f(x) \), the right-hand limit \( \lim_{x \to 0^+} f(x) \), and the value of the function at \( x = 0 \), which is \( f(0) = b \), must all be equal.

### Step 1: Left-hand limit as \( x \to 0^- \)

For \( x < 0 \), we have:

\[
f(x) = \frac{3x^2}{\sqrt{x^2 + a^2} - a}
\]

Taking the limit as \( x \to 0^- \):

\[
\lim_{x \to 0^-} f(x) = \frac{3(0)^2}{\sqrt{0^2 + a^2} - a} = 0
\]

So the left-hand limit is \( 0 \).

### Step 2: Right-hand limit as \( x \to 0^+ \)

For \( x > 0 \), we have:

\[
f(x) = \frac{b}{3^{1 - \tan x} + 3^{1 - \cot x}}
\]

Taking the limit as \( x \to 0^+ \), both \( \tan(0) = 0 \) and \( \cot(0) = \infty \), so:

\[
\lim_{x \to 0^+} f(x) = \frac{b}{3^{1 - 0} + 3^{1 - \infty}} = \frac{b}{3 + 0} = \frac{b}{3}
\]

So the right-hand limit is \( \frac{b}{3} \).

### Step 3: Continuity at \( x = 0 \)

For continuity at \( x = 0 \), the left-hand limit, right-hand limit, and \( f(0) = b \) must all be equal. Therefore, we set:

\[
\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)
\]

This gives us the following conditions:

1. \( \lim_{x \to 0^-} f(x) = 0 \) implies \( f(0) = b = 0 \).
2. \( \lim_{x \to 0^+} f(x) = \frac{b}{3} \), and since \( b = 0 \), this condition is also satisfied.

### Final Answer:

The constants are:
\[
a = 0, \quad b = 0
\]

Would you like further clarification or details? 

Here are 5 questions that build on this:
1. What is the definition of continuity for a piecewise function?
2. How does one compute limits of piecewise functions at a point of interest?
3. Why does setting \( a = 0 \) ensure the continuity for the left-hand limit?
4. What role does the value of \( b \) play in the right-hand limit?
5. How would the analysis change if the interval for the function were different?

**Tip**: For limits involving piecewise functions, always evaluate each piece separately and check continuity by comparing the limits at the boundary points.

A function is defined on the interval (-π/2, π/2) by a multipart rule. The constants a and b need to be determined to make the function continuous at x=0.

This problem explores the continuity of a piecewise function defined on the interval (-π/2, π/2). Learn how to find constants a and b to ensure that the function is continuous at x = 0, using limits and properties of trigonometric functions. Suitable for students in advanced calculus courses.

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