The minimum value of `4e ^(2x) + 9e^(-2x) ` is

Answer

To find the minimum value of the expression \( 4e^{2x} + 9e^{-2x} \), 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: \( a = 4e^{2x} \) and \( b = 9e^{-2x} \).

2. **Apply the 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:
\[ 4e^{2x} + 9e^{-2x} \geq 2\sqrt{(4e^{2x})(9e^{-2x})} \]

3. **Calculate the product**: Now, calculate \( ab \):
\[ ab = (4e^{2x})(9e^{-2x}) = 36 \]

4. **Find the square root**: Therefore,
\[ \sqrt{ab} = \sqrt{36} = 6 \]

5. **Substitute back into the inequality**: Now substituting back into the AM-GM inequality:
\[ 4e^{2x} + 9e^{-2x} \geq 2 \cdot 6 = 12 \]

6. **Conclusion**: The minimum value of \( 4e^{2x} + 9e^{-2x} \) is \( 12 \). This minimum value occurs when \( 4e^{2x} = 9e^{-2x} \).

### Verification of equality condition:
To find when equality holds in the AM-GM inequality, we set:
\[ 4e^{2x} = 9e^{-2x} \]
This leads to:
\[ 4e^{4x} = 9 \implies e^{4x} = \frac{9}{4} \implies 4x = \ln\left(\frac{9}{4}\right) \implies x = \frac{1}{4} \ln\left(\frac{9}{4}\right) \]
Thus, the minimum value of \( 4e^{2x} + 9e^{-2x} \) is indeed \( 12 \).