How do you identify all the critical points for $x^2 – x + y^2 + y = 0$ ?

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Answer 1 · 2016-11-02T18:48:52

Please see the explanation for a description of how one does it.

Use implicit differentiation, to find the first derivative, $(dy/dx)$ :

$2x -1 + 2y(dy/dx) + dy/dx = 0$

Separate $dy/dx$ from everything else

$2x -1 + (2y + 1)(dy/dx) = 0$

$(2y + 1)(dy/dx) = 1 - 2x$

$dy/dx = (1 - 2x)/(2y + 1)$

The critical points occur when the first derivative is equal to zero:

$0 = (1 - 2x)/(2y + 1)$

$0 = (1 - 2x)$

$2x = 1$

$x = 1/2$

Find the two values of y where this occurs:

$y^2 + y - 1/2 + 1/4 = 0$

$y^2 + y - 1/4 = 0$

check the discriminant:

$b^2 - 4(a)(c) = 1 - 4(1)(-1/4) = 2$

$y = -1/2 + sqrt(2)/2 and y = -1/2 - sqrt(2)/2$

The critical points are $(1/2,-1/2 + sqrt(2)/2)$ and $(1/2,-1/2 - sqrt(2)/2)$

How do you identify all the critical points for x^2 – x + y^2 + y = 0 x^2 – x + y^2 + y = 0?