Second derivative test. Worked example recitation
Critical points of functions, how to find them using the first derivatives, and how to classify 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 by the following formula: w = x cubed minus 3xy plus y cubed. We’re asked to find the critical values of this function and classify them as either minima or maxima or saddle points, using the second derivative test. To find critical points, we need to look at the first derivative. The critical points are where all partial derivatives or one partial derivative if there is more than one variable are 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 by evaluating the derivative at (x.y) = (0,0); this gives 3×2 – 3y2 = 0, which when simplified gives us x + y = 0. And we also want the derivative of our function with respect 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 first equation and solve for y in terms of x. That would give us y equals x squared. And then if we plug y equals x squared into the second equation, we get -3x plus 3x squared squared. So that’s x to the fourth equals 0, and we could divide out by that 3. 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 sides by x and then find that x cubed equals 1. Since there are no other solutions, x equals 0 or 1. The corresponding y values are obtained by squaring x. Thus, when x is 0, y 1; is 0; when x is 1, y is 1; so this function has two critical points, (0, 0), and (1, 1). to Now we need to determine whether the critical points are maxima, minima, saddle points or some combination of these. To accomplish this, we’ll use the second derivative test. In order to apply the second derivative test, one needs the second derivatives. So let’s compute them. We take our first partial, 3x squared minus 3y, and we take another partial of it with respect to X. 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 as yy. 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 W and w sub yx, which are equal to each other whenever our is function is nicely behaved, like polynomial. So w sub xy: we just take the two mixed partials, and so we take the partial of w sub x with respect to y; for example, 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 the equations. We look at the expression AC – B7, which is called the discriminant. We want to know whether this is positive or negative at the critical points-the points where two or more curves intersect. Let’s find the value of point (1, 1) first. At this point, we have A=6. We find that putting x=1 and y=1 into the expression for C gives us 6. Similarly, putting x=1 and y=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 or minimum value for the expression AC minus B squared. When this expression is positive, as it is in our case, we can conclude that there exists a maximum or minimum value. In order to find out whether we have a maximum or minimum, we must check the sign of A-in this case it’s positive. When the absolute value of B squared is positive and A is positive, you have a minimum. The critical point (1, 2) is a 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 was equal to 6y everywhere. At (0, 0), this simplifies to 6x times O equals 0, or 0 is constant. Likewise, y times O equals 0, or y is constant. So AC minus B squared is O times 0 minus 9 = -9. Because -9 is less than zero, this means that our function has a saddle point at (0, 0). So, in this case, the second derivative test was able to distinguish what kinds of critical points we had and found that the first critical point (1, 1) was a minimum and that the second critical point (0, 0) was a saddle point. Find and classify the critical points of the function w=x^3 -3xy+y^3 w_x= 3x^2-3y-0=>y=x^2 W_y=-3x +3y^2=0=>-x^4+x =0 x = 0 or 1 Critical points (0,0) (1,1)
