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 December 7th, 2012, 07:55 PM #1 Newbie   Joined: Dec 2012 Posts: 1 Thanks: 0 Implicit and critical numbers? the function is f(x) = x^2+xy +y^2-3=0 and you are told to differentiate it. I got dy/dx = (-2x-y)/(x+2y). I believe this is correct but the second part asks you to find where the function is "rising and falling", which I take to mean where it is increasing and decreasing. I know that youw ould usually set the derivated equal to zero and solve, but how would you do this for this problem since it has two variables?
December 7th, 2012, 08:10 PM   #2
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Re: Implicit and critical numbers?

Here is a plot of the curve along with the lines we get when we equate the numerator and denominator of the derivative to zero:

[attachment=0:3jmbyee6]slantedellipse.jpg[/attachment:3jmbyee6]

What does this tell us?
Attached Images
 slantedellipse.jpg (24.4 KB, 82 views)

 December 8th, 2012, 08:36 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Implicit and critical numbers? You know that $y'= -\frac{2x+ y}{x+ 2y}$ and you know the function will be increasing when that is positive, decreasing when it is negative. That is, this is a matter of determining when $-\frac{2x+y}{x+ 2y}> 0$ and when $-\frac{2x+y}{x+2y}< 0$. You should also know two things: a continuous function can only change sign when it is 0, and a rational function like this is 0 only when the numerator is 0 and discontinuous only when the denominator is 0. That means that this function can change sign only when 2x+y= 0, which gives y= -2x so that you can put into $x^2+ xy+ y^2= 3$ and solve for x, or when x+ 2y= 0 which occurs on the line y= -x/2, which can, again, put into $x^2+ xy+ y^2= 3$ and solve for x. Once you have found the values of x that satisfy those equations, you can check a single point in each interval between them to determine if the derivative is positive or negative on that interval. Fortunately, those give the same x-values as those where MarkFL's line crosses the ellipse!

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