Second Derivative Test. Worked Example Recitation

Second derivative test. Worked example recitation Critical points of functions, how to find them using the first derivatives, and how toclassify them using the second derivative test. So we have a function, w, that's a polynomial in two variables, x and y. It's given bythe following formula: w = x cubed minus 3xy plus y cubed. We're asked to find thecritical values of this function and classify them as either minima or maxima orsaddle points, using the second derivative test. To find critical points, we need to look at the first derivative. The critical points arewhere all partial derivatives or one partial derivative if there is more than one variableare equal to zero. We need to look at the first derivative. The first partial derivatives of a function with respect to x and y can be computed byevaluating the derivative at (x,y) = (0,0); this gives 3x2 – 3y2 = 0, which whensimplified gives us x + y = 0. And we also want the derivative of our function withrespect to y to equal 0; this gives us 3x – 3y2 = 0, which simplifies to –3x + 2y = 0. Solving these equations is fairly simple. We could, for example, take the firstequation and solve for y in terms of x. That would give us y equals x squared. Andthen if we plug y equals x squared into the second equation, we get -3x plus 3xsquared squared. So that's x to the fourth equals 0, and we could divide out by that3. So that means -x plus x to the fourth equals 0.We can solve this equation by first setting x equal to 0, or we can divide both sidesby x and then find that x cubed equals 1. Since there are no other solutions, x equals0 or 1. The corresponding y values are obtained by squaring x. Thus, when x is 0, yis 0; when x is 1, y is 1; so this function has two critical points, (0, 0), and (1, 1). Now we need to determine whether the critical points are maxima, minima, saddlepoints or some combination of these.To accomplish this, we'll use the second derivative test. In order to apply the secondderivative test, one needs the second derivatives. So let's compute them. We takeour first partial, 3x squared minus 3y, and we take another partial of it with respect tox. So in this case, that's just going to be 6x.Next, we have the second partial of y with respect to x, which can be expressed asyy. We then take a second partial of this expression, but this time with respect to y.That will simply be 6y.We have the mixed partials w sub xy and w sub yx, which are equal to each otherwhenever our function is nicely behaved, like a polynomial. So w sub xy, we just takethe two mixed partials, and so we take the partial of w sub x with respect to y, forexample, and that gives us minus 3. So these are our three partials.

Then we examine the expressions A, B, and C, which are common to each of theequations. We look at the expression AC – B², which is called the discriminant. Wewant to know whether this is positive or negative at the critical points—the pointswhere two or more curves intersect. Let's find the value of point (1, 1) first. At this point, we have A=6. We find thatputting x=1 and y=1 into the expression for C gives us 6. Similarly, putting x=1 andy=1 into the expression for B gives us −3. Thus, A is 6. The equation AC minus B squared equals 27 implies that there exists a maximum orminimum value for the expression AC minus B squared. When this expression ispositive, as it is in our case, we can conclude that there exists a maximum orminimum value. In order to find out whether we have a maximum or minimum, wemust check the sign of A—in this case it's positive. When the absolute value of Bsquared is positive and A is positive, you have a minimum. The critical point (1, 2) isa local minimum for this function. Now, we can take the same approach for the critical point (0, 0). As you may recall,A was equal to 6x everywhere, B was equal to negative 3 everywhere, and C wasequal to 6y everywhere. At (0, 0), this simplifies to 6x times 0 equals 0, or 0 isconstant. Likewise, y times 0 equals 0, or y is constant. So AC minus B squared is 0times 0 minus 9 = -9. Because -9 is less than zero, this means that our function hasa saddle point at (0, 0). So, in this case, the second derivative test was able to distinguish what kinds ofcritical points we had and found that the first critical point (1, 1) was a minimum andthat the second critical point (0, 0) was a saddle point.

So, to review what we did, we had a function, w, from here to here. We computed itsfirst derivatives and set them equal to zero. We solved that system of equations andfound two distinct solutions for that system of equations. Those solutions correspondto two critical points on the graph of this function, which we need to test with thesecond derivative test in order to determine whether they are saddles, minima ormaxima. We considered two critical points: (1, 1) and (0, 0). At those points weevaluated the second derivative. The first derivative at the point (1, 1) is A. Themixed partial derivative of w sub xy with respect to x at the point (1, 1) is B. The firstderivative at 0 is C. So we evaluate these expressions at the critical points and thenlook at AC minus B squared. If it's negative we have a saddle point; if it's positive weeither have a maximum or minimum; so we check which one by looking at the sign ofA.