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*(613*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: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Derivative problem
You have: $\displaystyle f(x)=x^{\frac{6}{7}}x^{\frac{13}{7}}$ $\displaystyle f \,'(x)=\frac{6}{7}x^{\frac{1}{7}}\frac{13}{7}x^{\frac{6}{7}}=\frac{1}{7}x^{\frac{1}{7}}\left(613x\right)=\frac{613x}{7x^{\frac{1}{7}}}$ 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: 12,211 Thanks: 521 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 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,957 Thanks: 1146 Math Focus: Elementary mathematics and beyond  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: 20,916 Thanks: 2199 
I've done some further patching. I suspect the original post contains a typing error.


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