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For example, for $f\left(x,y\right)=y^3+3x^2y-6x^2-6y^2+2$,

$f_x\left(a,b\right)=$ $6xy-12x$

and

$f_y\left(a,b\right)=$ $3y^2+3x^2-12y$

How many critical points am I looking for? How do I know when I've found them all?

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  • $\begingroup$ There are only two critical points. $\endgroup$ – user122283 May 6 '14 at 23:44
  • $\begingroup$ I assume you meant $f_x(x,y)$ and $f_y(x,y)$? $\endgroup$ – Hurkyl May 7 '14 at 0:02
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$\tag 1 f_x\left(x,y\right)=6xy-12x = 0, f_y\left(x,y\right)= 3y^2+3x^2-12y = 0$

From the first equation we have:

$$6xy-12x = 0 \implies 6x(y - 2) = 0 \implies x = 0, y = 2$$

We now use these in the second:

$$3y^2+3x^2-12y = 0$$

At $x = 0$, we have:

$$3y(y-4) = 0 \implies y = 0, y = 4$$

At $y = 2$, we have:

$$12+3x^2-24 = 0 \implies x^2 = 4 \implies x = \pm~ 2$$

Putting this all together, we have:

$$(x, y) = (0, 0), (0, 4), (-2, 2), (2,2)$$

Thus, we have a total of four critical points.

Note: sometimes things can get squirrely and it helps if you can draw an contour plot for each equation and see the number of intersection points. For example:

enter image description here

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Finding the critical points of one function, is equivalent to finding the zeroes of another function (its derivative).

There is no general rule for how many zeroes a function can have. $\sin x$ has infinitely many zeroes, while $\sin x+2$ has none.

But for polynomials, there can't be more zeroes than the order of the polynomial.

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You first invoke the theorem that tells you, in this case, every critical point is a solution to the system of equations $f_x = 0$ and $f_y = 0$.

Thus, if you find the complete solution to this system of equations, you have all of the critical points.

This is just a matter of high school algebra; the manipulations you've learned for solving problems do not "lose" any of the solutions. So, as long as you don't make mistakes like dividing by $x$ when you don't actually know that $x$ is nonzero, you will find all of the critical points.

However, the manipulations you've learned can create new, "spurious" solutions to the simplified problems that aren't solutions to the original problem. This is why you're supposed to check all of the solutions you find to be sure they really are solutions.

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