My Math Forum Derivative problem

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 March 19th, 2011, 09:10 PM #1 Newbie   Joined: Feb 2011 Posts: 21 Thanks: 0 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.
 March 20th, 2011, 11:04 AM #2 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 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.
 March 20th, 2011, 11:20 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs 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.
 March 20th, 2011, 02:36 PM #4 Newbie   Joined: Feb 2011 Posts: 21 Thanks: 0 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.
 March 20th, 2011, 03:15 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs 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.
 March 8th, 2018, 11:52 PM #6 Newbie   Joined: Feb 2018 From: Burnaby Posts: 5 Thanks: 0 Replies 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.
March 9th, 2018, 12:30 AM   #7
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Quote:
 Originally Posted by moody 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.

 March 9th, 2018, 02:36 AM #8 Newbie   Joined: Feb 2018 From: Burnaby Posts: 5 Thanks: 0 Thank you!!! Last edited by skipjack; March 9th, 2018 at 06:41 AM.
 March 9th, 2018, 07:44 AM #9 Global Moderator   Joined: Dec 2006 Posts: 19,522 Thanks: 1747 I've done some further patching. I suspect the original post contains a typing error.

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### or f(x)=x6/7−x13/7 find the critical numbers and open intervals where the function is increasing and decreasing.

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