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

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(B) Let $f(x) = 4e^{2x} + 9e^{-2x}$.To find the minimum value,we use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality.For any two positive

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Let $f(x) = 4e^{2x} + 9e^{-2x}$.To find the minimum value,we use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality.For any two positive numbers $a$ and $b$,$\frac{a+b}{2} \ge \sqrt{ab}$.Here,$a = 4e^{2x}$ and $b = 9e^{-2x}$.$\frac{4e^{2x} + 9e^{-2x}}{2} \ge \sqrt{4e^{2x} 	imes 9e^{-2x}}$$\frac{4e^{2x} + 9e^{-2x}}{2} \ge \sqrt{36 	imes e^{2x-2x}}$$\frac{4e^{2x} + 9e^{-2x}}{2} \ge \sqrt{36 	imes 1}$$\frac{4e^{2x} + 9e^{-2x}}{2} \ge 6$$4e^{2x} + 9e^{-2x} \ge 12$.Thus,the minimum value is $12$.

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

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