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 \( 4^x + 4^{2-x} \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Here’s a step-by-step solution:

### Step 1: Rewrite the Expression
The expression we want to minimize is:
\[ y = 4^x + 4^{2-x} \]

### Step 2: Apply AM-GM Inequality
According to the AM-GM inequality, for any non-negative numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
with equality when \( a = b \).

In our case, let:
\[ a = 4^x \quad \text{and} \quad b = 4^{2-x} \]

Then, we apply the AM-GM inequality:
\[ \frac{4^x + 4^{2-x}}{2} \geq \sqrt{4^x \cdot 4^{2-x}} \]

### Step 3: Simplify the Right Side
Calculating the right side:
\[ \sqrt{4^x \cdot 4^{2-x}} = \sqrt{4^{x + (2-x)}} = \sqrt{4^2} = \sqrt{16} = 4 \]

### Step 4: Set Up the Inequality
Now substituting back into our inequality:
\[ \frac{4^x + 4^{2-x}}{2} \geq 4 \]
Multiplying both sides by 2 gives:
\[ 4^x + 4^{2-x} \geq 8 \]

### Step 5: Determine When Equality Holds
The equality in the AM-GM inequality holds when \( a = b \). Thus, we set:
\[ 4^x = 4^{2-x} \]
Taking logarithm base 4 of both sides:
\[ x = 2 - x \]
Solving for \( x \):
\[ 2x = 2 \implies x = 1 \]

### Step 6: Find the Minimum Value
Substituting \( x = 1 \) back into the original expression:
\[ 4^1 + 4^{2-1} = 4 + 4 = 8 \]

### Conclusion
Thus, the minimum value of \( 4^x + 4^{2-x} \) is:
\[ \boxed{8} \]

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