Find the minimum value of $4e^{2x} + 9e^{-2x}$.

12

The correct answer is Option (3) → 12 ##

Let $f(x) = 4e^{2x} + 9e^{-2x}$

$∴f'(x) = 8e^{2x} - 18e^{-2x}$

Put $f'(x) = 0$

$\Rightarrow 8e^{2x} - 18e^{-2x} = 0$$

$\Rightarrow e^{4x} = \frac{18}{8} = \frac{9}{4} \Rightarrow e^{2x} = \frac{3}{2} \Rightarrow x = \log\left(\frac{3}{2}\right)^{1/2}$

Again $f''(x) = 16e^{2x} + 36e^{-2x} > 0$

$= 6 + 6 = 12$$

Now, $f\left(\log\left(\frac{3}{2}\right)^{1/2}\right) = 4e^{2\log\left(\frac{3}{2}\right)^{1/2}} + 9e^{-2\log\left(\frac{3}{2}\right)^{1/2}}$

$= 4 \times \frac{3}{2} + 9 \times \frac{2}{3}$

$=6+6$

$=12$