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Topic review (newest first)

bob bundy
2012-09-26 04:42:31

hi n1corponic

You are welcome.  Post again if you need to.  smile

Bob

n1corponic
2012-09-26 04:33:06

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.

smile

Bob

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

bob bundy
2012-09-26 04:25:55

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

smile

Bob

n1corponic
2012-09-26 04:13:35

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! big_smile 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?

bob bundy
2012-09-26 04:03:33

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

n1corponic
2012-09-26 03:56:59

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??

bobbym
2012-09-26 03:54:30
bob bundy
2012-09-26 03:49:05

My post 4 may have been missed as bobbym and I posted together.

Bob

n1corponic
2012-09-26 03:45:12

bobbym wrote:

Hi;

Did you look at a graph at all?

Hi bobbym! smile 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.

bob bundy
2012-09-26 03:44:22

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

bobbym
2012-09-26 03:37:52

Hi;

Did you look at a graph at all?

bob bundy
2012-09-26 03:35:32

hi n1corponic

I'm on the case.  Stay on-line.

Welcome!

Bob

n1corponic
2012-09-26 03:17:42

...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.. hmm

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