The minimum value of $4e^{2x} + 9e^{-2x}$ is
A11
B12
C10
D14
Answer & Solution
Correct answer: B. 12
Let $t=e^{2x}$. Then $t>0$, and the expression becomes $$4t+\frac{9}{t}.$$ By AM-GM, $$4t+\frac{9}{t}\ge 2\sqrt{4t\cdot \frac{9}{t}}=2\sqrt{36}=12.$$ Equality holds when $$4t=\frac{9}{t}.$$ This gives $$4t^2=9,$$ so $$t=\frac{3}{2},$$ which is possible since $t>0$. Hence the minimum value is $12$. Comparing with the options, this matches option $\text{B}$.
Related questions
The angle made by the tangent of the curve $x = a$ ($t + in t$ cost); $y = a(1 + in t)^2A value of $C$ for which the conclusion of mean value theorem holds for the function $f(x)The value of $f(0)$ so that the function $f(x) = \frac{ qrt{1 + x} - qrt[3]{1 + x}}{x}$ b$\int_{-2}^{2}|1 - x^2|dx =$$\lim_{n\to \infty} um_{r = 1}^{n}1 - \frac{1}{n} e^{r}$ isLet $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$, for all $x \in [0, If $\frac{dv}{dx} = y + 3 > 0$ and $y(0) = 2$, then $y(\ln 2)$ is equal toThe equation $y^{2} \mathrm{e}^{xy} = 9 \mathrm{e}^{-3} \cdot x^{2}$ defines $y$ as a diff