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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The minimum value of the function $f(x) = x \log x$ is

If $ax + \frac{b}{x} \ge c$ for all positive $x$,where $a, b > 0$,then:

$A(1,15), B(3,-12), C(6,12)$ are three consecutive turning points of a continuous curve $y=f(x)$. If $f(x)=0$ only for $x=\alpha$ and $x=\beta$,then $|\beta-\alpha| < $

Find the absolute maximum and minimum values of the function $f$ given by $f(x) = \cos^{2} x + \sin x$,where $x \in [0, \pi]$.

It is given that at $x=1,$ the function $f(x) = x^{4}-62x^{2}+ax+9$ attains its maximum value on the interval $[0,2].$ Find the value of $a.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module