You are not logged in.
#1 2012-09-26 03:17:42
- n1corponic
- Novice
Offline
Can somebody please help me understand...
...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.. 
#3 2012-09-26 03:37:52
- bobbym
- Administrator
Offline
Re: Can somebody please help me understand...
Hi;
Did you look at a graph at all?
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
#4 2012-09-26 03:44:22
- bob bundy
- Moderator
Offline
Re: Can somebody please help me understand...
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
- Novice
Offline
Re: Can somebody please help me understand...
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.
#7 2012-09-26 03:54:30
- bobbym
- Administrator
Offline
Re: Can somebody please help me understand...
Hi;
This might help too.
http://ltcconline.net/greenl/courses/11 … crease.htm
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
#8 2012-09-26 03:56:59
- n1corponic
- Novice
Offline
Re: Can somebody please help me understand...
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
- Moderator
Offline
Re: Can somebody please help me understand...
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
- Novice
Offline
Re: Can somebody please help me understand...
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
- Moderator
Offline
Re: Can somebody please help me understand...
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
- Novice
Offline
Re: Can somebody please help me understand...
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!