The minimum value of `4^(x)+4^(2-x),x in R` is

Options

A. 0
B. 2
C. 4
D. 8

Answer

To find the minimum value of the expression \( 4^x + 4^{2-x} \) for \( x \in \mathbb{R} \), we can use the concept of the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Identify the terms**: Let \( A = 4^x \) and \( B = 4^{2-x} \). We want to find the minimum value of \( A + B \).

2. **Apply AM-GM Inequality**: According to the AM-GM inequality, we have:
\[ \frac{A + B}{2} \geq \sqrt{AB} \]
This implies:
\[ A + B \geq 2\sqrt{AB} \]

3. **Calculate \( AB \)**:
\[ AB = 4^x \cdot 4^{2-x} = 4^{x + (2-x)} = 4^2 = 16 \]

4. **Substitute \( AB \) into the AM-GM Inequality**:
\[ A + B \geq 2\sqrt{16} = 2 \cdot 4 = 8 \]

5. **Conclusion**: Therefore, we have shown that:
\[ 4^x + 4^{2-x} \geq 8 \]
The minimum value of \( 4^x + 4^{2-x} \) is \( 8 \).

6. **Finding when equality holds**: The equality in the AM-GM inequality holds when \( A = B \). Thus,
\[ 4^x = 4^{2-x} \]
Taking logarithm (base 4) on both sides gives:
\[ x = 2 - x \implies 2x = 2 \implies x = 1 \]
Therefore, the minimum value occurs at \( x = 1 \).

### Final Answer:
The minimum value of \( 4^x + 4^{2-x} \) is \( \boxed{8} \).