Finding the least value of $y=3^{x-1}+3^{-x-1}$

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.

• Do you mean $3y=x+\frac 1x?$
– mfl
Jul 18, 2017 at 15:55
• Yes, you are right @mfl Jul 18, 2017 at 15:56
• The way this is written is very confusing to read, please could you exapnd on the method you used to attempt to solve this Jul 18, 2017 at 15:56
• 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 $x=1.$
– mfl
Jul 18, 2017 at 15:57

we have by AM-GM: $$\frac{3^{x-1}+3^{-x-1}}{2}\geq \sqrt{3^{x-1}\cdot 3^{-x-1}}=\frac{1}{3}$$
$y' = 3^{x-1}\ln 3 + 3^{-x-1}\ln 3\cdot (-1)$. 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 $$y'<0 \iff 3^{-x} > 3^{x} \iff x<0$$