The expansion of `(3x+2)^(-1/2)` is valid in ascending powers of x if x lies in the interval.

Answer

To find the interval in which the expansion of \((3x + 2)^{-1/2}\) is valid in ascending powers of \(x\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (3x + 2)^{-1/2} \]
To facilitate the expansion, we can factor out the constant term from the expression:
\[ (3x + 2)^{-1/2} = \left(2\left(\frac{3x}{2} + 1\right)\right)^{-1/2} \]
This simplifies to:
\[ = 2^{-1/2} \left(1 + \frac{3x}{2}\right)^{-1/2} \]

### Step 2: Identify the condition for the binomial expansion
The binomial expansion \((1 + u)^{n}\) is valid for \(|u| < 1\). Here, we have:
\[ u = \frac{3x}{2} \]
Thus, the condition for the expansion to be valid is:
\[ \left| \frac{3x}{2} \right| < 1 \]

### Step 3: Solve the inequality
We can rewrite the inequality:
\[ -1 < \frac{3x}{2} < 1 \]
Now, we will solve the two parts of this compound inequality.

#### Part 1: Solve \(-1 < \frac{3x}{2}\)
Multiplying through by 2 gives:
\[ -2 < 3x \]
Dividing by 3 results in:
\[ -\frac{2}{3} < x \]

#### Part 2: Solve \(\frac{3x}{2} < 1\)
Similarly, multiplying through by 2 gives:
\[ 3x < 2 \]
Dividing by 3 results in:
\[ x < \frac{2}{3} \]

### Step 4: Combine the results
Combining both parts of the inequality, we have:
\[ -\frac{2}{3} < x < \frac{2}{3} \]

### Conclusion
Thus, the interval in which the expansion of \((3x + 2)^{-1/2}\) is valid in ascending powers of \(x\) is:
\[ x \in \left(-\frac{2}{3}, \frac{2}{3}\right) \]

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