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**n1corponic****Member**- Registered: 2012-09-10
- Posts: 7

...how we find the interval where the function e^x-e^(3x) is increasing... the derivative is e^x[1 - 3e^(2x)]. I have the answer but I feel lost..

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi n1corponic

I'm on the case. Stay on-line.

Welcome!

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,654

Hi;

Did you look at a graph at all?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bob bundy****Moderator**- Registered: 2010-06-20
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OK, here we go.

Your differentiation is correct. For an increasing function you want dy/dx to be > 0

e^x is always positive so you need

log base e is an increasing function so the inequality holds if you take logs

Graph below.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**n1corponic****Member**- Registered: 2012-09-10
- Posts: 7

bobbym wrote:

Hi;

Did you look at a graph at all?

Hi bobbym! I took a look at the graph in wolframalpha or smth..but it won't help on my future exams.. if i do not know how to find the answer on my own.

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**bob bundy****Moderator**- Registered: 2010-06-20
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My post 4 may have been missed as bobbym and I posted together.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,654

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**n1corponic****Member**- Registered: 2012-09-10
- Posts: 7

bob bundy wrote:

OK, here we go.

Your differentiation is correct. For an increasing function you want dy/dx to be > 0

e^x is always positive so you need

log base e is an increasing function so the inequality holds if you take logs

Graph below.

Bob

Your approach seems wonderful!!! and according to me correct.. but why does the book i have give another..but very similar answer?.. it says x<or=-1/2*ln3 Is it the same but i'm missing smth??

*Last edited by n1corponic (2012-09-25 05:59:02)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi n1corponic

They are the same because

But the book shouldn't be saying less than and equal to because the graph has a turning point when = so dy/dx = 0

So the value of the function isn't getting bigger.

Bob

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**n1corponic****Member**- Registered: 2012-09-10
- Posts: 7

bob bundy wrote:

hi n1corponic

They are the same because

But the book shouldn't be saying less than and equal to because the graph has a turning point when = so dy/dx = 0

So the value of the function isn't getting bigger.

Bob

wow! they are the same indeed! I feel sooooo much better now!

Well, the book says (-infinity, -1/2*ln3] , so I guess it includes -1/2*ln3 probably because the question doesn't ask where the fuction strictly...increases..that must be it! right?

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**bob bundy****Moderator**- Registered: 2010-06-20
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The square bracket means include the endpoint. It seems I am assuming too much. If it doesn't say strictly, I concede the argument.

Bob

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**n1corponic****Member**- Registered: 2012-09-10
- Posts: 7

bob bundy wrote:

The square bracket means include the endpoint. It seems I am assuming too much. If it doesn't say strictly, I concede the argument.

Bob

Thank you for everything!!!!!!! bob bundy!!!!!! plus congrats to owner..admins..moderators etc. for this great site! cheers!

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi n1corponic

You are welcome. Post again if you need to.

Bob

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