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

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(B) Let $f(x) = 4e^{2x} + 9e^{-2x}$.Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive real numbers $a$ and $b$,we have $\

Vedclass · Accepted Answer

$12$

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(B) Let $f(x) = 4e^{2x} + 9e^{-2x}$.Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive real numbers $a$ and $b$,we have $\frac{a+b}{2} \ge \sqrt{ab}$,which implies $a+b \ge 2\sqrt{ab}$.Here,let $a = 4e^{2x}$ and $b = 9e^{-2x}$.Then $f(x) = 4e^{2x} + 9e^{-2x} \ge 2\sqrt{(4e^{2x})(9e^{-2x})}$.$f(x) \ge 2\sqrt{36e^{2x-2x}} = 2\sqrt{36e^0} = 2 	imes 6 = 12$.The equality holds when $4e^{2x} = 9e^{-2x}$,i.e.,$e^{4x} = \frac{9}{4}$,which is possible for some real $x$.Thus,the minimum value is $12$.

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

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