What is the domain of the function
f(x) log `sqrt(2x^(2) - 15)` ?

Options

A. `-7.5 lt x lt 7.5`
B. `x lt -7.5 or x gt 7.5`
C. `x lt -2.7 or x gt 2.7`
D. `x lt -3.2 or x gt 3.2`

Explanation

To find the domain of the function \( f(x) = \log(\sqrt{2x^2 - 15}) \), we need to ensure that the expression inside the logarithm is positive. The logarithmic function is defined only for positive arguments.

### Step 1: Set the argument of the logarithm greater than zero
We start by setting the expression inside the logarithm greater than zero:
\[ \sqrt{2x^2 - 15} > 0 \]

### Step 2: Square both sides
Since the square root is always non-negative, we can square both sides without changing the inequality:
\[ 2x^2 - 15 > 0 \]

### Step 3: Solve the inequality
Now, we solve the inequality:
\[ 2x^2 > 15 \]
Dividing both sides by 2, we get:
\[ x^2 > \frac{15}{2} \]
\[ x^2 > 7.5 \]

### Step 4: Take the square root
Next, we take the square root of both sides. Remember that taking the square root introduces both positive and negative solutions:
\[ x > \sqrt{7.5} \quad \text{or} \quad x < -\sqrt{7.5} \]

### Step 5: Approximate the square root
Calculating \(\sqrt{7.5}\):
\[ \sqrt{7.5} \approx 2.74 \]
Thus, we have:
\[ x > 2.74 \quad \text{or} \quad x < -2.74 \]

### Conclusion: Domain of the function
The domain of the function \( f(x) = \log(\sqrt{2x^2 - 15}) \) is:
\[ (-\infty, -2.74) \cup (2.74, \infty) \]