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.
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 $$y'<0 \iff 3^{-x} > 3^{x} \iff x<0$$
