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A question on derivative for real valued function

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Hi all, I have a probably simple question on a general function, let say I have a real valued function y = f(x), x >= 0

Now I have calculated that for all x, the 1st derivative is always negative and the 2nd derivative is always positive. With this finding, can I infer that, the 1st derivative will always decrease (in absolute value) as I move onward in x?

Thanks for your comment.

Thanks and regards,

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Consider a line with a negative slope

The first derivative will always be negative, so f(x) will always decrease.

But the absolute value of that line won't decrease forever. It'll decrease until it hits the x axis, and then abs(f(x)) will begin to increase (as f(x) become more negative)

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[quote name='ronm' timestamp='1336504589' post='4938463']
Now I have calculated that for all x, the 1st derivative is always negative and the 2nd derivative is always positive. With this finding, can I infer that, the 1st derivative will always decrease (in absolute value) as I move onward in x?
[/quote]

Yes.

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[quote name='ronm' timestamp='1336504589' post='4938463']
Now I have calculated that for all x, the 1st derivative is always negative and the 2nd derivative is always positive. With this finding, can I infer that, the 1st derivative will always decrease (in absolute value) as I move onward in x?
[/quote]

If the 1st derivative is always negative, but increasing, it would imply to me that the 1st derivative is asymptotic to 0 (or a value below 0), because that's the only way it could never become positive while having a positive 2nd derivative.

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[quote name='jefferytitan' timestamp='1336514806' post='4938500']
[quote name='ronm' timestamp='1336504589' post='4938463']
Now I have calculated that for all x, the 1st derivative is always negative and the 2nd derivative is always positive. With this finding, can I infer that, the 1st derivative will always decrease (in absolute value) as I move onward in x?
[/quote]

If the 1st derivative is always negative, but increasing, it would imply to me that the 1st derivative is asymptotic to 0 (or a value below 0), because that's the only way it could never become positive while having a positive 2nd derivative.
[/quote]

It doesn't have to be asymptotic to zero:
f(x) = 1/x - x
f'(x) = -1/x^2 - 1 < 0
f"(x) = 1/x^3 > 0
limi(f'(x)) = -1 as x->infinity Edited by alvaro

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@alvaro, if you read more carefully "[color=#282828][font=helvetica, arial, verdana, tahoma, sans-serif][size=3][left][background=rgb(250, 251, 252)]it would imply to me that the 1st derivative is asymptotic to 0 (or a value below 0)". e.g. asymptotic to z where z<=0.[/background][/left][/size][/font][/color]

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[quote name='jefferytitan' timestamp='1336543816' post='4938587']
@alvaro, if you read more carefully "
[left][background=rgb(250, 251, 252)]it would imply to me that the 1st derivative is asymptotic to 0 (or a value below 0)". e.g. asymptotic to z where z<=0.[/background][/left]

[/quote]
The text in parentheses is supposed to be explanatory, and if I ignore it the sentence shouldn't become false. I was just clarifying that the value can indeed be below 0.

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