Minimum value at `a^(ax) + a^(1-a x); a gt 0` and `x in R`, is

Options

A. `sqrta`
B. `a/sqrt2`
C. `2sqrta`
D. `1/(2sqrta)`

Answer

To find the minimum value of the expression \( a^{ax} + a^{1 - ax} \) where \( a > 0 \) and \( x \in \mathbb{R} \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Identify the Expression**:
We start with the expression:
\[ f(x) = a^{ax} + a^{1 - ax} \]

2. **Apply AM-GM Inequality**:
According to the AM-GM inequality, for any non-negative numbers \( A \) and \( B \):
\[ \frac{A + B}{2} \geq \sqrt{AB} \]
We can apply this to our expression by letting:
\[ A = a^{ax} \quad \text{and} \quad B = a^{1 - ax} \]
Thus, we have:
\[ \frac{a^{ax} + a^{1 - ax}}{2} \geq \sqrt{a^{ax} \cdot a^{1 - ax}} \]

3. **Simplify the Right-Hand Side**:
The product \( a^{ax} \cdot a^{1 - ax} \) can be simplified:
\[ a^{ax} \cdot a^{1 - ax} = a^{ax + (1 - ax)} = a^1 = a \]
Therefore, we can write:
\[ \frac{a^{ax} + a^{1 - ax}}{2} \geq \sqrt{a} \]

4. **Multiply Both Sides by 2**:
To eliminate the fraction, multiply both sides by 2:
\[ a^{ax} + a^{1 - ax} \geq 2\sqrt{a} \]

5. **Conclusion**:
The minimum value of \( a^{ax} + a^{1 - ax} \) occurs when \( a^{ax} = a^{1 - ax} \), which happens when \( ax = 1 - ax \) or \( 2ax = 1 \) leading to \( x = \frac{1}{2a} \). Therefore, the minimum value of the expression is:
\[ \boxed{2\sqrt{a}} \]