$\displaystyle e^{y}+e^{-y}\leq 2$

I am struggling to solve this inequality.

I do know $\displaystyle [0,x] \ x\in\mathbb{R}$

From guessing and checking, I know $\displaystyle \displaystyle x<\frac{1}{10}$

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- Dec 31st 2010, 05:23 PMdwsmithe^{y}+e^{-y}\leq 2
$\displaystyle e^{y}+e^{-y}\leq 2$

I am struggling to solve this inequality.

I do know $\displaystyle [0,x] \ x\in\mathbb{R}$

From guessing and checking, I know $\displaystyle \displaystyle x<\frac{1}{10}$ - Dec 31st 2010, 05:44 PMmr fantastic
First draw a graph of $\displaystyle w = e^{y}+e^{-y}$ - it has a turning point at (0, 2) and the shape

*appears*parabolic (it's not a parabola of course, it's a cosh function).

So your job is simply to solve $\displaystyle 2 = e^{y}+e^{-y}$:

$\displaystyle 2 = e^{y}+e^{-y}$

$\displaystyle \Rightarrow 2 e^y = (e^{y})^2 + 1$

$\displaystyle \Rightarrow (e^{y})^2 - 2 e^y + 1 = 0$

Solve this quadratic for $\displaystyle e^y$ (reject one of the solutions for obvious reasons) and hence solve for y. Now use the graph you drew to solve the given inequality. - Dec 31st 2010, 08:21 PMProve It
- Dec 31st 2010, 08:59 PMdwsmith
$\displaystyle e^{2y}-2e^y+1=(e^y-1)^2=0$

So the answer is just 0? - Jan 1st 2011, 12:17 AMProve It
- Jan 1st 2011, 02:26 AMHallsofIvy
Since $\displaystyle e^{y}$ is positive for all y, multiplying $\displaystyle e^y+ e^{-y}\le 2$ by $\displaystyle e^y$ gives $\displaystyle e^{2y}+ 1\le 2e^{y}$ or $\displaystyle e^{2y}- 2e^y+ 1= (e^y- 1)^2\le 0$. Since a square is never negative, that inequality is satisfied only when $\displaystyle (e^y- 1)= 0$ or when y= 0.

- Jan 1st 2011, 03:14 AMmr fantastic
- Jan 1st 2011, 07:32 AMchisigma
The inequality can be written as...

$\displaystyle \cosh y \le 1$ (1)

... and if we intend to find the values of $\displaystyle y$ so that $\displaystyle \cosh y$ is*real*and $\displaystyle \le 1$, then any $\displaystyle y= 0 + i\ w$ , $\displaystyle w \in \mathbb{R}$ satisfies (1)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$