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## #1 2012-09-26 03:17:42

n1corponic
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...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..

## #2 2012-09-26 03:35:32

bob bundy
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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

## #3 2012-09-26 03:37:52

bobbym

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Hi;

Did you look at a graph at all?

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #4 2012-09-26 03:44:22

bob bundy
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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

## #5 2012-09-26 03:45:12

n1corponic
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#### 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.

## #6 2012-09-26 03:49:05

bob bundy
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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

## #7 2012-09-26 03:54:30

bobbym

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In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #8 2012-09-26 03:56:59

n1corponic
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#### 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 (2012-09-26 03:59:02)

## #9 2012-09-26 04:03:33

bob bundy
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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

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

## #10 2012-09-26 04:13:35

n1corponic
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#### 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?

## #11 2012-09-26 04:25:55

bob bundy
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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

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

## #12 2012-09-26 04:33:06

n1corponic
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#### 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!

bob bundy
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