The domain of the function
`f(x)=root(6)(4^(x)+8^(2//3(x-2))-52-2^(2(x-1)))` is
(a)(0,1) (b)`[3, infty]` (c)[1,0) (d)None of these

Options

A. (0,1)
B. `[3, infty]`
C. [1,0)
D. None of these

Answer

To find the domain of the function

\[ f(x) = \sqrt[6]{4^x + 8^{\frac{2}{3}(x-2)} - 52 - 2^{2(x-1)}} \]

we need to ensure that the expression inside the sixth root is greater than or equal to zero:

\[ 4^x + 8^{\frac{2}{3}(x-2)} - 52 - 2^{2(x-1)} \geq 0 \]

### Step 1: Rewrite the expression

First, we can express \(8^{\frac{2}{3}(x-2)}\) in terms of base 2:

\[ 8 = 2^3 \implies 8^{\frac{2}{3}(x-2)} = (2^3)^{\frac{2}{3}(x-2)} = 2^{2(x-2)} = 2^{2x - 4} \]

Now, rewrite the function:

\[ f(x) = 4^x + 2^{2x - 4} - 52 - 2^{2(x-1)} \]

Next, we can express \(4^x\) as \(2^{2x}\):

\[ f(x) = 2^{2x} + 2^{2x - 4} - 52 - 2^{2x - 2} \]

### Step 2: Combine like terms

Now, let's combine the terms involving \(2^{2x}\):

\[ f(x) = 2^{2x} + \frac{1}{4} \cdot 2^{2x} - 52 - \frac{1}{4} \cdot 2^{2x} \]

This simplifies to:

\[ f(x) = 2^{2x} \left(1 + \frac{1}{4} - \frac{1}{4}\right) - 52 \]

This further simplifies to:

\[ f(x) = 2^{2x} - 52 \]

### Step 3: Set up the inequality

Now we need to solve the inequality:

\[ 2^{2x} - 52 \geq 0 \]

This can be rearranged to:

\[ 2^{2x} \geq 52 \]

### Step 4: Solve for \(x\)

Taking logarithm base 2 on both sides:

\[ 2x \geq \log_2(52) \]

Thus,

\[ x \geq \frac{1}{2} \log_2(52) \]

Calculating \(\log_2(52)\):

\[ \log_2(52) = \log_2(4 \cdot 13) = \log_2(4) + \log_2(13) = 2 + \log_2(13) \]

Using the approximation \(\log_2(13) \approx 3.7\):

\[ \log_2(52) \approx 2 + 3.7 = 5.7 \]

Thus,

\[ x \geq \frac{5.7}{2} \approx 2.85 \]

### Step 5: Find the domain

Since \(x\) must be greater than or equal to approximately 2.85, we can conclude that the domain of the function is:

\[ x \in [3, \infty) \]

### Final Answer

Thus, the correct option is:

(b) \([3, \infty)\)