You didn't complete the calculation. Yes, y= x or y= -x. Now put those back into the equations for z_x or z_y. With y= x, z_x= 2xy+ 1+y^2= 0 becomes 2x^2+ 1+ x^2= 0 or 3x^2= -1. That's impossible and gives no solution. With y= -x, z_x= 2xy+ 1+ y^2= 0 becomes -2x^2+ 1+ x^2= 0 or x^2= 1. That gives x= 1 or -1 and, since y= -x, the points are (1, -1) and (-1, 1).

and, for example, if h=0, k> 0, this is k(k)= k^2> 0. If h> 0, k= 0, it is h(-h)= -h^2< 0.I go on

evaluating the general point with co-ordinates x = 1+h and y = -1+k in the neighbourhood of (1,-1,0) I want to find the value of z and determining the nature of said point =>

z = (1+h-1+k)[(1+h)(-1+k)+1] = (h+k)(-1+k-h+hk+1) = (h+k)(k-h+hk)

Did you consider using the "second derivative test" for this? While it requires calculating the second derivative, it only requires evaluating at a single point so, here, is easier to use.similarly for the other S.P. => z = (h+k)(h-k+hk)

the book says these are saddle points and I also think that there is nothing to ensure that the z is for all small enough values of h and k either smaller or greater than zero, but I am reluctant and this shows more in

ii. z = x^2 + y + (2/x) + (4/y)

as before z_x = 2x – [2/(x^2)]

and z_y = 1 – [4/(y^2)]

and so the S.P.’s can be found (no problem) to be (1,2,7) and (1,-2,-1))

as before z = (1+h)^2 + (2+k) + [2/(1+h)] + [4/(2+k)]

and for the other S.P. z = (1+h)^2 + (-2+k) + [2/(1+h)] + [4/(-2+k)]

Now the book says that (1,2,7) is a minimum and (1,-2,-1) is a saddle point, but I can’t see this…?

If , evaluated at the critical point, is positive, then the point is either a maximum or a minimum. It is a minimum if f_{xx}> 0 (or f_{yy}> 0; they must have the same sign in this situation) and a maximum if f_{xx}< 0. If is negative, the point is a saddle point.

Here, , and .

At (1, 2) is (6)(1)- 0= 6> 0 while [tex]f_{xx}= 6> 0 so (1, 2) is a minimum.

At (1, -2) is (-6)(1)- 0= -6< 0 so (1, -2) is a saddlepoint.

Again, it is simpler to use the "second derivative test". , and .iii. z = xy + ln|x| + 2y^2

as before z_x = y + 1/x

and z_y = x + 4y

and so the S.P.’s can be found (no problem) to be (2, -1/2, ln2-1/2) and (-2, 1/2, ln2-1/2)

as before z= (2+h)(k – 1/2) + ln|2+h| + 2(k – 1/2)^2

and for the other S.P. z = (-2+h)(k + 1/2) + ln|-2+h| + 2(k + 1/2)^2

the book says that these are both saddle points and me I also think that the second and third terms must be positive but the first can be either negative or positive for small enough values of h and k, though here again I have the feeling that i’m in the wrong. I should be comparing the value of z to ln2-1/2 not zero but that seems even more confusing…

pls help me I am stydying on my own and will feel quite incompetent if I can’t get a better grasp of these stuff.

At (2, -1/2), so (2, -1/2) is a saddle point.

At (-2, 1/2) so (-2, 1/2) is also a saddle point.