Results 1 to 5 of 5

Thread: Find x-coordinate of stationary point

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Find x-coordinate of stationary point

    Find the x-coordinate of the stationary point of the curve $\displaystyle y=\frac{(x-1)^3}{x+1}$, $\displaystyle x>0$
    What I did

    $\displaystyle \frac{dy}{dx}=\frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}$

    Stationary Point, dy/dx=0

    $\displaystyle \frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}=0$
    $\displaystyle (x+1)(3)(x-1)^2-(x-1)^3=0$
    $\displaystyle 3(x+1)(x-1)^2=(x-1)^3$
    $\displaystyle 3(x+1)=(x-1)$
    $\displaystyle 3x+3=x-1$
    $\displaystyle 2x+4=0$
    $\displaystyle x=-2$

    however question states x>0 and ans is 1!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2010
    Posts
    186
    Thanks
    13
    You've made a mistake when you divided the expression $\displaystyle 3(x+1)(x-1)^2=(x-1)^3$ with the term $\displaystyle (x-1)^2$.

    This is what you should do:
    $\displaystyle (x+1)(3)(x-1)^2-(x-1)^3=0$
    $\displaystyle (x-1)^2(3(x+1)-(x-1))=0$
    $\displaystyle (x-1)^2(2x+4))=0$
    $\displaystyle 2(x-1)^2(x+2))=0$

    And now you see that the equality will hold if either $\displaystyle x-1=0$ or $\displaystyle x+2=0$, that is if $\displaystyle x=1$ or $\displaystyle x=-2.$ Since you have a constraint $\displaystyle x>0$ the correct answer is $\displaystyle x=1.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Thank you! but why was that a mistake?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2010
    Posts
    186
    Thanks
    13
    Because you cannot be sure whether or not you carried out a division by zero! For an example, if you want to divide something with an expression (x-1) than you should point out that you exclude the possibility of x taking the value 1, thus you loose a possible solution, which exactly was the case here. And I suspect that wast the idea of the given problem, to see if one is aware of what I just said in the first two sentences of this post.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003
    Quote Originally Posted by Punch View Post
    Thank you! but why was that a mistake?
    As Mathoman said, you cannot divide by 0. What you could have done, rather than factor, was, at the point where you had
    $\displaystyle 3(x+ 1)(x- 1)^2= (x- 1)^3$
    was think "If x- 1 is not 0, I can divide both sides by $\displaystyle (x- 1)^2$ to get 3(x+ 1)= x- 1, 3x+ 3= x- 1, 2x= -4, x= -2. But if x- 1= 0, then both sides would equal 0 so x= 1 is also a solution."

    If the condition "x> 0" had not been included, both x= -2 and x= 1 would have given stationary points. With that condition, only x= 1 does.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find x coordinate of a point on a line
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: Feb 12th 2011, 10:42 PM
  2. Find the x−coordinate of the point A
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Oct 7th 2010, 05:58 AM
  3. Stationary point...
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Jun 9th 2010, 06:57 PM
  4. stationary point
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 2nd 2009, 09:24 PM
  5. stationary point
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Nov 16th 2008, 01:45 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum