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Topic: Help Needed in math
Replies: 25   Last Post: May 2, 2014 6:01 PM

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William Elliot

Posts: 1,669
Registered: 1/8/12
Re: Help Needed in math
Posted: May 2, 2014 12:21 AM
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On Thu, 1 May 2014, quasi wrote:
> >>>>
> >>>>f(x) = (5x + 3) / x^2, x <> 0

> >>>What's f?
> >Did you define f as f(x) = (5x - 3)/x^2?
>
> No, he wrote
> f(x) = (5x + 3)/x^2
>

> >>>>If a < f(x) for all x, then a has to be less than the
> >>>>global minimum of f(x).
> >>>>The global minimum of f(x) = -25/12.
> >>>>The largest integer less than this is -3.


f'(x) = 5/x^2 - 2(5x + 3)/x^3 = 0; -5x - 6 = 0; x = -6/5

The extremum point, x = -6/5 when f(x) = -25/12, is a minimum.

> Not when you use the correct f'(x).
>

> >What you're doing is unfathomable.
> Unfathomable to you perhaps.
> Nevertheless, konyberg's argument is correct.

His answer is correct. I don't see his reasoning
How does f(x) relate to the missing problem?




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