If x is a real number,what is the minimum val | vedclass

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(A) Given the function $f(x) = 3^{x+1} + 3^{-(x+1)}$.We can rewrite this as $f(x) = 3^{x+1} + \frac{1}{3^{x+1}}$.Since $3^{x+1} > 0$ for all real $x$,

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given the function $f(x) = 3^{x+1} + 3^{-(x+1)}$.We can rewrite this as $f(x) = 3^{x+1} + \frac{1}{3^{x+1}}$.Since $3^{x+1} > 0$ for all real $x$,we can apply the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.For any two positive numbers $a$ and $b$,$\frac{a+b}{2} \geq \sqrt{ab}$,which implies $a+b \geq 2\sqrt{ab}$.Let $a = 3^{x+1}$ and $b = \frac{1}{3^{x+1}}$.Then $f(x) = a + b \geq 2\sqrt{a \cdot b}$.$f(x) \geq 2\sqrt{3^{x+1} \cdot \frac{1}{3^{x+1}}}$.$f(x) \geq 2\sqrt{1}$.$f(x) \geq 2$.Thus,the minimum value of the function is $2$.

If $x$ is a real number,what is the minimum value of $f(x) = 3^{x+1} + 3^{-(x+1)}$?

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