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Hey guys. We are working on functions: increasing, decreasing, monotone, bounded, minimums, and maximums.
Last edited by sumpm1 (2009-03-07 02:37:23)
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Please do 1(a) and 1(b) yourself. They are straightforward enough.
I will help you with part of 1(c) because that one might be tricky.
Now do the same for
.Last edited by JaneFairfax (2009-03-08 01:17:40)
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Jane, you need to stop this. You don't yell at a person for asking a question, even for something they should know. If you do that, people stop asking all questions, including the good ones. Instead, you need to say something along the lines of, "What are you having problems with?" I really don't want to have to say this again, please tone it down.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Just use the definitions of the two that you provided.
For part c I'll give you a hint that;
sinh(2x) = 2sinh(x)cosh(x)
cosh(2x) = cosh^2(x) + sinh^2(x)
Again use the definition of cosh(x) and sinh(x) and manipulate the numbers to get their identities.
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Hey guys, thanks for the help.
@ksmathwiz: Are you sure that the identity for cosh(2x) is not cosh(2x) = cosh^2(x) - sinh^2(x) rather than cosh(2x) = cosh^2(x) + sinh^2(x)?
Thanks
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ksmathwizs formulas are correct.
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Here: (e^2x + e^-2x )/2
= (2e^2x + 2e^-2x)/4
= (e^2x + 2 + e^-2x + e^2x 2 + e^-2x)/4
You finish it.
Last edited by ksmathwiz (2009-03-07 04:37:21)
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We can prove that
is increasing by contradiction.Suppse it is not. Then there exist
in I such that .Now
and .Also
is either or .Thus we have either
or . This contradicts the fact that both f and g are increasing on I.Therefore
is increasing on I.You can prove that
is also increasing on the interval I by the same method.Last edited by Muggleton (2009-03-07 07:42:47)
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Incidentally:
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