Find the domain of definitions of the following function: `f(x)=sqrt(3-2^(x)-2^(1-x))`

Explanation

To find the domain of the function \( f(x) = \sqrt{3 - 2^x - 2^{1-x}} \), we need to ensure that the expression under the square root is non-negative. This means we need to solve the inequality:

\[ 3 - 2^x - 2^{1-x} \geq 0 \]

### Step 1: Rewrite the inequality
We can rewrite \( 2^{1-x} \) as \( \frac{2}{2^x} \). Thus, the inequality becomes:

\[ 3 - 2^x - \frac{2}{2^x} \geq 0 \]

### Step 2: Multiply through by \( 2^x \)
To eliminate the fraction, we can multiply the entire inequality by \( 2^x \) (noting that \( 2^x > 0 \) for all real \( x \)):

\[ 3 \cdot 2^x - 2^{2x} - 2 \geq 0 \]

### Step 3: Rearranging the inequality
Rearranging gives us:

\[ -2^{2x} + 3 \cdot 2^x - 2 \geq 0 \]

### Step 4: Let \( t = 2^x \)
Now, let \( t = 2^x \). The inequality becomes:

\[ - t^2 + 3t - 2 \geq 0 \]

### Step 5: Multiply by -1
Multiplying the entire inequality by -1 (which reverses the inequality) gives:

\[ t^2 - 3t + 2 \leq 0 \]

### Step 6: Factor the quadratic
Factoring the quadratic, we have:

\[ (t - 1)(t - 2) \leq 0 \]

### Step 7: Determine the intervals
The critical points are \( t = 1 \) and \( t = 2 \). We can test the intervals:

1. For \( t < 1 \): Choose \( t = 0 \): \( (0 - 1)(0 - 2) = 2 > 0 \)
2. For \( 1 < t < 2 \): Choose \( t = 1.5 \): \( (1.5 - 1)(1.5 - 2) = (-0.25) < 0 \)
3. For \( t > 2 \): Choose \( t = 3 \): \( (3 - 1)(3 - 2) = 2 > 0 \)

Thus, the solution to the inequality \( (t - 1)(t - 2) \leq 0 \) is:

\[ 1 \leq t \leq 2 \]

### Step 8: Substitute back for \( t \)
Recall that \( t = 2^x \). Therefore, we have:

\[ 1 \leq 2^x \leq 2 \]

### Step 9: Solve for \( x \)
Taking logarithms, we find:

1. From \( 2^x \geq 1 \): \( x \geq 0 \)
2. From \( 2^x \leq 2 \): \( x \leq 1 \)

### Conclusion
Thus, the domain of the function \( f(x) \) is:

\[ [0, 1] \]