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

**dwsmith** · December 31st 2010, 05:23 PM

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

I am struggling to solve this inequality.

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

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

**mr fantastic** · December 31st 2010, 05:44 PM

Originally Posted by dwsmith

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

I am struggling to solve this inequality.

I do know $[0,x] \ x\in\mathbb{R}$ Mr F says: Why are you using y above but x here and below?
From guessing and checking, I know $\displaystyle x<\frac{1}{10}$

First draw a graph of $w = e^{y}+e^{-y}$ - 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 $2 = e^{y}+e^{-y}$ :

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

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

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

Solve this quadratic for $e^y$ (reject one of the solutions for obvious reasons) and hence solve for y. Now use the graph you drew to solve the given inequality.

**Prove It** · December 31st 2010, 08:21 PM

Originally Posted by mr fantastic

First draw a graph of $w = e^{y}+e^{-y}$ - 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 $2 = e^{y}+e^{-y}$ :

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

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

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

Solve this quadratic for $e^y$ (reject one of the solutions for obvious reasons) and hence solve for y. Now use the graph you drew to solve the given inequality.

Adding to Mr F's very useful post, now you have a quadratic inequality in $\displaystyle e^y$ . Quadratic inequalities are easiest solved by completing the square.

**dwsmith** · December 31st 2010, 08:59 PM

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

So the answer is just 0?

**Prove It** · January 1st 2011, 12:17 AM

Originally Posted by dwsmith

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

So the answer is just 0?

Remember that you are trying to solve $\displaystyle e^{2y} - 2e^y + 1 \leq 0$ , not $\displaystyle e^{2y} - 2ey + 1 = 0$ ...

And yes, the answer is 0...

**HallsofIvy** · January 1st 2011, 02:26 AM

Since $e^{y}$ is positive for all y, multiplying $e^y+ e^{-y}\le 2$ by $e^y$ gives $e^{2y}+ 1\le 2e^{y}$ or $e^{2y}- 2e^y+ 1= (e^y- 1)^2\le 0$ . Since a square is never negative, that inequality is satisfied only when $(e^y- 1)= 0$ or when y= 0.

**mr fantastic** · January 1st 2011, 03:14 AM

Originally Posted by dwsmith

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

So the answer is just 0?

Yes. And is in fact easily seen from the graph without having to do the algebra.

**chisigma** · January 1st 2011, 07:32 AM

The inequality can be written as...

$\cosh y \le 1$ (1)

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

Kind regards

$\chi$ $\sigma$

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