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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. 
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.
Bob
Thank you for everything!!!!!!! bob bundy!!!!!! plus congrats to owner..admins..moderators etc. for this great site! 
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.
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! 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
Hi;
This might help too.
http://ltcconline.net/greenl/courses/11 … crease.htm
- 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! 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.. 