The minimum value of `9^x+9^(1-x),x inR,` is

Options

A. 2
B. 3
C. 4
D. none of these.

Explanation

To find the minimum value of the expression \(9^x + 9^{1-x}\) where \(x \in \mathbb{R}\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Identify the Terms**:
We have two terms in the expression: \(a = 9^x\) and \(b = 9^{1-x}\).

2. **Apply AM-GM Inequality**:
According to the AM-GM inequality, for any non-negative real numbers \(a\) and \(b\):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
This implies:
\[ 9^x + 9^{1-x} \geq 2\sqrt{9^x \cdot 9^{1-x}} \]

3. **Calculate the Geometric Mean**:
We can simplify the geometric mean:
\[ ab = 9^x \cdot 9^{1-x} = 9^{x + (1-x)} = 9^1 = 9 \]
Therefore, the geometric mean is:
\[ \sqrt{ab} = \sqrt{9} = 3 \]

4. **Substituting Back into AM-GM**:
Now substituting back into the AM-GM inequality:
\[ \frac{9^x + 9^{1-x}}{2} \geq 3 \]
Multiplying both sides by 2 gives:
\[ 9^x + 9^{1-x} \geq 6 \]

5. **Finding the Minimum Value**:
The minimum value of \(9^x + 9^{1-x}\) occurs when the equality condition of the AM-GM inequality holds, which is when \(a = b\). This occurs when:
\[ 9^x = 9^{1-x} \]
Taking logarithm on both sides, we find:
\[ x = 1 - x \implies 2x = 1 \implies x = \frac{1}{2} \]

6. **Calculating the Minimum Value**:
Substituting \(x = \frac{1}{2}\) back into the original expression:
\[ 9^{1/2} + 9^{1 - 1/2} = 3 + 3 = 6 \]

### Conclusion:
Thus, the minimum value of \(9^x + 9^{1-x}\) is \(6\).