Stationary points of curve y = x/(1+2x)^2

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Monkeyseat

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Question

A curve has equation y = x/(1+2x)^2, find the coodinates of the stationary points.

Working

Using the quotient rule:

u = x
du/dx = 1

v = (1 + 2x)^2
dv/dx = 4 + 8x

dy/dx = ((1 + 2x)^2 - x(4 + 8x))/(1 + 2x)^4
dy/dx = ((1 + 2x)^2 - (4x + 8x^2))/(1 + 2x)^4
dy/dx = (1/(1 + 2x)^2) - (4x/(1 + 2x)^3)

After simplifying a bit more:

dy/dx = (1 - 2x)/(1 + 2x)^3

Stationary point when dy/dx = 0:

(1 - 2x)/(1 + 2x)^3 = 0
1 - 2x = 0
2x = 1
x = 1/2

When x = 1/2, y = 1/8.

That's one stationary point, but the question implies that there is more than one... How can I find the other one?

I initially solved ((1 + 2x)^2 - (4x + 8x^2))/(1 + 2x)^4 = 0 without simplifying and found that x could equal -1/2 also, but the denominator comes out as 0 when you sub back in for y so that can't be right.

Many thanks.

Sorry if the layout is a bit confusing.
 
A stationary point is where f'(x)=0. You seem to have nailed it. I see no more than that.
 
If there is only one stationary point then I guess it must just be a misprint in the question. It's pretty annoying, because I was sat here for ages trying to figure out where I went wrong.

Thanks for clarifying galactus.
 
Sometimes the books just print it that way because many of the problems do have more than 1. Do not take it so literal.

Here is a graph. We can see it there at x=1/2. If we look at the numerator of the derivative, we see it is 1-2x. The only thing that makes that 0 is x=1/2
 

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Ah, i'll be aware of that for future questions. Thanks again. :)

The only thing that confused me was about what happens to the curve at x = -0.5. When I sketched it, the two sides of the 'dip' on negative side of the graph seemed to touch. After closer inspection, I have found that they do not, so the graph is not defined for x = -0.5
 
The function has two critical points (x = ± 0.5) - however one stationary point (x = 0.5).

Sometimes these two properties get mixed up.
 
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