
#1
August 6th, 2014,
10:39

August 6th, 2014 10:39
# ADS
Circuit advertisement

MHB Journeyman
#2
August 6th, 2014,
10:43
On the last integral, is the upper bound "9"?

#3
August 6th, 2014,
10:44
Thread Author
Originally Posted by
Rido12
On the last integral, is the upper bound "9"?
Yes.

Pessimist Singularitarian
#4
August 6th, 2014,
10:44
Hello and welcome to MHB!
Can you tell us what you've tried or what your thoughts are on how to begin, so our helpers have an idea where you are stuck and then can provide better help?

#5
August 6th, 2014,
10:47
Thread Author
Originally Posted by
MarkFL
Hello and welcome to MHB!
Can you tell us what you've tried or what your thoughts are on how to begin, so our helpers have an idea where you are stuck and then can provide better help?
Thank you so much!
I began with the condition that the secant must lie above the curve at all points if the curve is concave upward but that seem to lead nowhere. I also tried to utilise the fact that the double derivative of f(X) will be positive but to no avail.
Any help will be appreciated.

Pessimist Singularitarian
#6
August 6th, 2014,
11:00
I think I would begin with the minimal case:
$ \displaystyle 0<f^{\prime\prime}(x)$
So, integrating, what can you say about $f$?

#7
August 6th, 2014,
11:03
Thread Author
Originally Posted by
MarkFL
I think I would begin with the minimal case:
$ \displaystyle 0<f^{\prime\prime}(x)$
So, integrating, what can you say about $f$?
Wouldn't integrating this inequality conclude the same thing as what we are given, that f is a concave upward function?

Pessimist Singularitarian
#8
August 6th, 2014,
11:14
Originally Posted by
mathisfun
Wouldn't integrating this inequality conclude the same thing as what we are given, that f is a concave upward function?
What did you get when integrating?

#9
August 6th, 2014,
11:19
Thread Author
Originally Posted by
MarkFL
What did you get when integrating?
That f(x) is greater than zero, but that seems to be contradictory as a negative function may also have positive derivative.

Pessimist Singularitarian
#10
August 6th, 2014,
11:42
Let's begin with:
$ \displaystyle 0<f^{\prime\prime}(x)$
And so, on $ \displaystyle [1,k]$ where $ \displaystyle 1<k\in\mathbb{R}$ there must be some constant $C$ such that:
$ \displaystyle C<f^{\prime}(x)$
What happens if you integrate again?