Problem 608

Let $f(x) = e^{x+1} + e^{-x} - 2$. Because $f(x)$ approaches infinity as $|x|$ increases without bound, the only possible axis of symmetry is a vertical line. If the axis of symmetry has equation $x = c$, then $f(x) = f(2c-x)$ for every real $x$, which is equivalent to $e \cdot e^x + e^{-x} = e^{2c+1}e^{-x} + e^{-2c}e^x$. Multiplying through by $e^x$ and simplifying gives $(e - e^{-2c})e^{2x} = e^{2c+1} - 1$. Because this equation holds for all $x$, it follows that $e - e^{-2c} = 0$ and $e^{2c+1} - 1 = 0$. Thus $c = -\frac{1}{2}$. The reflection of $(-1, \frac{1}{2})$ with respect to the vertical line $x = -\frac{1}{2}$ is $(0, \frac{1}{2})$.