I am struggling to solve this inequality.

I do know

From guessing and checking, I know

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

I am struggling to solve this inequality.

I do know

From guessing and checking, I know - December 31st 2010, 05:44 PMmr fantastic
First draw a graph of - 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 :

Solve this quadratic for (reject one of the solutions for obvious reasons) and hence solve for y. Now use the graph you drew to solve the given inequality. - December 31st 2010, 08:21 PMProve It
- December 31st 2010, 08:59 PMdwsmith

So the answer is just 0? - January 1st 2011, 12:17 AMProve It
- January 1st 2011, 02:26 AMHallsofIvy
Since is positive for all y, multiplying by gives or . Since a square is never negative, that inequality is satisfied only when or when y= 0.

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

(1)

... and if we intend to find the values of so that is*real*and , then any , satisfies (1)...

Kind regards