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  4. The minimum value of 4e2x+9e-2x is

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

Application of Derivatives

Solution:

Let $f (x) = 4e^{2x}+9e^{-2x}$
$\therefore f' (x) = 8e^{2x}-18e^{-2x}$
put $f'(x) =0$
$\Rightarrow 8e^{2x}-18e^{-2x}=0$
$e^{2x}=3/2$
$\Rightarrow x=log (3/2)^{1/2}$
Again $f'' (x) =16e^{2x}+36e^{-2x}>\,0$
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 $=12$