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March 19th, 2011, 09:10 PM   #1
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Derivative problem

Formula is f(x) = x^(6/7) - x^(13/7)


I got f'(x) = (1/7*(6-13*x)x^(1/7). Is it correct?

Also, f(x)=0 if x=___ ( I got 6/13, is it correct?) and that f'( ____) (I think it's x) does not exist.

Therefore, the critical numbers are____ and_____.

Last edited by skipjack; March 9th, 2018 at 06:49 AM.
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March 20th, 2011, 11:04 AM   #2
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Re: Derivative problem

Hello,
is this your function:
$\displaystyle x^{(6/7)} - x^{(13/7)}$?
if it is ( and I strongly think so )
then your answer should have been:
$\displaystyle (6/7)*x^{-(1/7)} - (13/7)*x^{(6/7)}$

$\displaystyle 2. f(x)=0, \text{ if } x=0 ,\, x=1$

critical point is at x=(6/13)^(1/7) which means the seventh
root of (6/13)
best regards al7aq

Last edited by skipjack; March 9th, 2018 at 07:41 AM.
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March 20th, 2011, 11:20 AM   #3
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Re: Derivative problem

You have:

$\displaystyle f(x)=x^{\frac{6}{7}}-x^{\frac{13}{7}}$

$\displaystyle f
\,'(x)=\frac{6}{7}x^{-\frac{1}{7}}-\frac{13}{7}x^{\frac{6}{7}}=\frac{1}{7}x^{-\frac{1}{7}}\left(6-13x\right)=\frac{6-13x}{7x^{\frac{1}{7}}}$

Thus:

$\displaystyle f\,'\!\left(\frac{6}{13}\right)=0$ and $\displaystyle f\,'(0)$ does not exist.

Therefore the critical numbers are $\displaystyle 0,\,6/13$.

Last edited by skipjack; March 9th, 2018 at 07:16 AM.
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March 20th, 2011, 02:36 PM   #4
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Re: Derivative problem

Thanks. I also got these questions and I'm wondering whether my answers are right.

f is increasing on an interval if f'(x)___ (my answer: >0) for each x on that interval. Therefore the given function is increasing on the interval ( ___, ___ ) (my answer: 0, 6/13)

f is decreasing on an interval if f'(x)____ (my answer:<0) for each x on that interval. Therefore the given function is decreasing on (–infinity), ___ (my answer:6/13)), ( ___(my answer:6/13), infinity).

Last edited by skipjack; March 9th, 2018 at 07:34 AM.
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March 20th, 2011, 03:15 PM   #5
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Re: Derivative problem

We check the three open intervals:

$\displaystyle \left(-\infty,0\right)$ we can use -1: f'(-1) is negative, f(x) is decreasing on this interval.

$\displaystyle \left(0,\frac{6}{13}\right)$ we can use �: f'(�) is positive, f(x) is increasing on this interval.

$\displaystyle \left(\frac{6}{13},\infty\right)$ we can use 1: f'(1) is negative, f(x) is decreasing on this interval.

Last edited by greg1313; March 9th, 2018 at 12:26 AM.
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March 8th, 2018, 11:52 PM   #6
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Hi, I can never see the answers posted - there's always a question mark; is there something wrong with my account?

Last edited by skipjack; March 9th, 2018 at 06:40 AM.
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March 9th, 2018, 12:30 AM   #7
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Quote:
Originally Posted by moody View Post
Hi, I can never see the answers posted - there's always a question mark; is there something wrong with my account?
No. Often with these old posts the format is not compatible with the newer software we are running. I've edited as much as possible so the posts are now more accessible.

Last edited by skipjack; March 9th, 2018 at 06:41 AM.
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March 9th, 2018, 02:36 AM   #8
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Thank you!!!

Last edited by skipjack; March 9th, 2018 at 06:41 AM.
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March 9th, 2018, 07:44 AM   #9
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I've done some further patching. I suspect the original post contains a typing error.
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