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- Thread starter mathisfun
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- Jul 18, 2013

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On the last integral, is the upper bound "9"?

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Yes.On the last integral, is the upper bound "9"?

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Thank you so much!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?

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.

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Wouldn't integrating this inequality conclude the same thing as what we are given, that f is a concave upward function?I think I would begin with the minimal case:

\(\displaystyle 0<f^{\prime\prime}(x)\)

So, integrating, what can you say about $f$?

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What did you get when integrating?Wouldn't integrating this inequality conclude the same thing as what we are given, that f is a concave upward function?

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That f(x) is greater than zero, but that seems to be contradictory as a negative function may also have positive derivative.What did you get when integrating?

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\(\displaystyle 0<f^{\prime\prime}(x)\)

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?

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If f1(X)>f2(x)

\(\displaystyle 0<f^{\prime\prime}(x)\)

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?

doesn't implies that the indefinite integration of f1(x) will be greater than f2(x). This is only applicable when both sides are integrated within limits and thus when integrating from 0 to x, constant gets eliminated.

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