# Find x-coordinate of stationary point

• Sep 13th 2010, 11:31 PM
Punch
Find x-coordinate of stationary point
Find the x-coordinate of the stationary point of the curve $y=\frac{(x-1)^3}{x+1}$, $x>0$
What I did

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

Stationary Point, dy/dx=0

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

however question states x>0 and ans is 1!
• Sep 13th 2010, 11:46 PM
MathoMan
You've made a mistake when you divided the expression $3(x+1)(x-1)^2=(x-1)^3$ with the term $(x-1)^2$.

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

And now you see that the equality will hold if either $x-1=0$ or $x+2=0$, that is if $x=1$ or $x=-2.$ Since you have a constraint $x>0$ the correct answer is $x=1.$
• Sep 13th 2010, 11:47 PM
Punch
Thank you! but why was that a mistake?
• Sep 14th 2010, 12:13 AM
MathoMan
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.
• Sep 14th 2010, 03:24 AM
HallsofIvy
Quote:

Originally Posted by Punch
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
$3(x+ 1)(x- 1)^2= (x- 1)^3$
was think "If x- 1 is not 0, I can divide both sides by $(x- 1)^2$ 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.