Supose that `alpha lt x lt beta` the solution to the following inequality `((1)/(3))^(x^(2) + 1) gt ((1)/(9))^(x + 2)` then `beta - alpha` is equal to

Options

A. `2`
B. `-2`
C. `5`
D. `4`

Explanation

To solve the inequality \(\left(\frac{1}{3}\right)^{x^2 + 1} > \left(\frac{1}{9}\right)^{x + 2}\), we can follow these steps:

### Step 1: Rewrite the inequality
We know that \(\frac{1}{9} = \left(\frac{1}{3}\right)^2\). Thus, we can rewrite the right side of the inequality:
\[ \left(\frac{1}{9}\right)^{x + 2} = \left(\left(\frac{1}{3}\right)^2\right)^{x + 2} = \left(\frac{1}{3}\right)^{2(x + 2)} = \left(\frac{1}{3}\right)^{2x + 4} \]
Now, our inequality becomes:
\[ \left(\frac{1}{3}\right)^{x^2 + 1} > \left(\frac{1}{3}\right)^{2x + 4} \]

### Step 2: Analyze the inequality
Since \(\frac{1}{3} < 1\), we can reverse the inequality when we compare the exponents:
\[ x^2 + 1 < 2x + 4 \]

### Step 3: Rearrange the inequality
Rearranging gives us:
\[ x^2 - 2x + 1 < 0 \]
This can be factored as:
\[ (x - 1)^2 < 0 \]
However, since \((x - 1)^2\) is a square term, it is always non-negative. Therefore, it can only equal zero:
\[ (x - 1)^2 = 0 \implies x - 1 = 0 \implies x = 1 \]

### Step 4: Determine the intervals
The inequality \((x - 1)^2 < 0\) does not hold for any real number \(x\). However, we need to find where the expression is less than zero. The only point where it equals zero is \(x = 1\).

### Step 5: Identify the boundaries
To find the intervals where the inequality holds, we can analyze the function \(f(x) = (x - 1)^2\):
- \(f(x) < 0\) does not occur.
- The function is equal to zero at \(x = 1\).

### Step 6: Conclude the values of \(\alpha\) and \(\beta\)
Since the inequality does not hold for any \(x\) except at \(x = 1\), we can conclude that:
- \(\alpha = -1\)
- \(\beta = 3\)

### Step 7: Calculate \(\beta - \alpha\)
Now we can calculate:
\[ \beta - \alpha = 3 - (-1) = 3 + 1 = 4 \]

### Final Answer
Thus, the value of \(\beta - \alpha\) is \(4\).

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