Question

The minimum value of $4e^{2x}+9e^{-2x}$ is

Answer 1

Answer: 12

Let $f(x)=4e^{2x}+9e^{-2x}$
$\therefore f^{\prime }(x)=8e^{2x}-18e^{-2x}$
put $f^{\prime }(x)=0$
$\Rightarrow 8e^{2x}-18e^{-2x}=0$
$e^{2x}=3/2$
$\Rightarrow x=log(3/2{)}^{1/2}$
Again $f^{\prime \prime }(x)=16e^{2x}+36e^{-2x}>0$
Now $f\left(log{\left(3/2\right)}^{1/2}\right)=4e^{2\left(log{\left(3/2\right)}^{1/2}\right)}+9e^{-2\left(log{\left(3/2\right)}^{1/2}\right)}$
$=4\times \frac{3}{2}+9\times \frac{2}{3}=6+6=12$
Hence, minimum value $=12$