Finding the lowest value possible of $f(x) = e^{1-0.5x}+e^{0.5x-1}$ 
Let $f$ be the function defined by: $$f(x) = e^{1-0.5x}+e^{0.5x-1}$$
Find the values of $k$ for which the equation $f(x) = k$ has two non
  negative solutions, using the graphical calculator.

This problem could be solved very easily using the calculator, but I want to try solving solving it analitically.
If you look at the graph of this equation http://www.wolframalpha.com/input/?i=e%5E%7B1-0.5x%7D%2Be%5E%7B0.5x-1%7D you'll notice that the wanted values are in the set ]lowest value of $f(x)$; $f(0)$]
Finding $f(0)$ is easy:
$$ e^{1-0.5\cdot 0}+e^{0.5\cdot 0-1} = e + e^{-1}$$
I think there are 2 ways of solving it:


*

*Simplify this equation to the form $ax^2 +bx +c$, and so c would be the wanted value;

*Solve $e^{1-0.5\cdot x}+e^{0.5\cdot x-1} < e + e^{-1} $, or $f(x) < f(0)$
For the first option I tried:
$$e^{1-0.5\cdot x}+e^{0.5\cdot x-1} = e^{1-\frac{x}{2}}+e^{\frac{x}{2}-1} = \sqrt{e^{2-x}} + \sqrt{e^{x-2}} = \frac{e^2+e^x}{e^{\frac{x}{2}+1}} = ???$$
For the second one I tried:
$$e^{1-0.5\cdot x}+e^{0.5\cdot x-1} < e + e^{-1} \\ \Leftrightarrow \frac{e}{e^{\frac{x}{2}}} + \frac{e^{\frac{x}{2}}}{e} < e + e^{-1}  \\ \Leftrightarrow  \frac{1}{e^{\frac{x}{2}}} \cdot e + \frac{1}{e} \cdot e^{\frac{x}{2}}  < e + e^{-1}$$
Then I did $y = e^{\frac{x}{2}}$
$$\frac{1}{y} \cdot e + y \cdot \frac{1}{e}  < e + e^{-1} \Leftrightarrow ????$$
My questions:


*

*How do I solve this using method 1 (if it is possible)?

*How do I solve this using method 2? 

*Can I factorize $\frac{1}{y} \cdot e + y \cdot \frac{1}{e}$? How?
Thanks
 A: Hint: let $a = e^{1-0.5x}$, then obviously $a \gt 0$ and $f(x) = a + \cfrac{1}{a} \ge 2$ by AM-GM, with equality iff $a=1$. Therefore the minimum of $f(x)$ is $2$ and is attained for $a=1 \iff e^{1-0.5x} = 1$.
A: For the second way:
\begin{equation}
\frac{1}{y} \cdot e + y \cdot \frac{1}{e}  < e + e^{-1}\\\frac{1}{y} \cdot e + y \cdot \frac{1}{e}  - e - e^{-1}<0\\\frac{e^2+y^2-e^2y-y}{ey}<0
\end{equation}
Now we know that the denominator is always positive, so we have to solve $e^2+y^2-e^2y-y<0$. This is a parabola thus it s easy:
\begin{equation}
e^2+y^2-e^2y-y=0\\y^2+y(-e^2-1)+e^2=0\\y=\frac{e^2+1\pm(e^2-1)}{2}
\end{equation}
hence we get $y=1$ and $y=e^2$. Thus the solution of the inequality is $1< y< e^2$ but $y=e^{x\over2}$ so $1< e^{x\over2}< e^2\longrightarrow 0<x<4$.
A: $$y = e^{1 - 0.5x} + e^{-(1 - 0.5x)}$$
Multiplying both sides by $e^{1 - 0.5x}$:
$$ye^{1 - 0.5x} = e^{2(1 - 0.5x)} + 1$$
Substitution for x makes it more clear what we're doing here:
$$yu = u^2 + 1$$
$$(u - \frac y2)^2 + 1 - \frac{y^2}4 = 0$$
The vertex happens when $1 - \frac{y^2}4 = 0$, or when $y = ±2$. Since we know this, that means $u = ±1$. Substituting back:
$$e^{1 - 0.5x} = ±1$$
Getting -1 for a real x is impossible. Letting $u = 1$ forces $y = 2$:
$$1 - 0.5x = 0 \rightarrow x = 2$$
Thus the minimum is at $(2, 2)$. Any value $k > 2$ gives two solutions.
A: Set $y=e^{0.25x-0.5}$. Then your expression is
$$
y^2+y^{-2}=2+(y-y^{-1})^2
$$
which is minimal exactly when $y=y^{-1}$ or $1=y^2=e^{0.5x-1}$, thus for $x=2$.
