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If $y=3^{x-1}+3^{-x-1}$ where $x$ is real, then what is the least value of $y$?

What I did: $$y=3^{x-1}+3^{-x-1}=3^x3^{-1}+3^{-x}3^{-1}$$ Let $3^x=z$, then $$y=\frac z3+\frac1{3z}$$ $$3y=z+\frac1z$$ Now what to do next? Is there any better way to solve this one

This is a gmat exam question.

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  • $\begingroup$ Do you mean $3y=x+\frac 1x?$ $\endgroup$
    – mfl
    Jul 18, 2017 at 15:55
  • $\begingroup$ Yes, you are right @mfl $\endgroup$
    – mathematics
    Jul 18, 2017 at 15:56
  • $\begingroup$ The way this is written is very confusing to read, please could you exapnd on the method you used to attempt to solve this $\endgroup$
    – lioness99a
    Jul 18, 2017 at 15:56
  • $\begingroup$ Fist of all $x>0$ (I suggest to change its name because you are using $x=3^x.$) Now, from the AM-GM inequality it is $x+\frac 1x\ge 2$ with equality iff $x=1.$ $\endgroup$
    – mfl
    Jul 18, 2017 at 15:57

2 Answers 2

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we have by AM-GM: $$\frac{3^{x-1}+3^{-x-1}}{2}\geq \sqrt{3^{x-1}\cdot 3^{-x-1}}=\frac{1}{3}$$

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$y' = 3^{x-1}\ln 3 + 3^{-x-1}\ln 3\cdot (-1)$. To find the minimum of $y$ we'll set the derivative to $0$ and find the critical points. We see $$y'=0 \iff 3^{x}=3^{-x}$$ Therefore the sole critical point occurs at $x=0$, and so it's $(0,y(0)) = \left(0, \frac{2}{3}\right)$. To verify the critical value is a minimum, observe that $$y'>0 \iff 3^{x} > 3^{-x} \iff x > 0$$ and $$y'<0 \iff 3^{-x} > 3^{x} \iff x<0$$

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