The Second Derivative Test   

Clearly, f( x,y) has a local maximum at a critical point ( p,q) only if every vertical slice of z = f(x,y) has a maximum at ( p,q) .

Similarly, f( x,y) has a local minimum at a critical point ( p,q) only if every vertical slice of z = f( x,y) has a minimum at ( p,q) .

However, it is possible for f( x,y) to have a minimum in one slice and a maximum in another slice.

If this is the case, then we say that f( x,y) has a saddle at ( p,q) , because the resulting surface resembles a saddle for a horse.

To determine if we get a maximum, a minimum, or a saddle point at a critical point ( p,q), we consider the vertical slice z(t) = f(p+mt, q+nt). Since x = p+mt and y = q+nt implies that x' (t) = m and y' (t) = n, the first derivative of z(t) is

dz dt = ¶f ¶x  dx dt + ¶f ¶y  dy dt = mfx+nfy

Moreover, m and n constant implies that

z" =   d dt dz dt = m  dfx dt +n  dfy dt = m æ è ¶fx ¶x  dx dt + ¶fx ¶y  dy dt ö ø +n æ è ¶fy ¶x  dx dt + ¶fy ¶y  dy dt ö ø = m( mfxx+nfxy ) +n( mfyx+nfyy )

Expanding and using the equality of the mixed partial derivatives then yields

z"(0) = m2fxx( p,q) + 2mnfxy( p,q) + n2fyy( p,q)

(2)

If f_xx(p,q) = 0, then we can choose values of m and n such that z''  is negative in some slices and positive in others, thus implying that z=f(x,y) has a saddle at (p,q). If f_xx(p,q) ¹ 0, then completing the square in m yields

z"(0) = fxx( p,q) æ è m +   fxy( p,q) fxx( p,q) n ö ø 2    +   D( p,q) fxx( p,q)  n2

(3)

where D = f_xxf_yy - ( f_xy) ²  is called the discriminant of f. (i.e., expanding (3) will result in (2) ).

If D( p,q) > 0, then z"(0) has the same sign as f_xx( p,q) in all directions u = <m,n>, thus implying a maximum if  f_xx( p,q) < 0 and a minimum if f_xx( p,q) > 0. However, if D( p,q) < 0, then choosing m=1, n=0 yields z"(0) > 0, whereas choosing m = f_xx( p,q) / f_xx( p,q) n yields  z"(0) < 0, thus implying a saddle.  These observations lead to the following theorem:           

Second Derivative Test: If ( p,q) is a critical point of a function f( x,y) whose second derivatives exist at ( p,q) , then

Discriminant 2nd der Result D( p,q) > 0, fxx( p,q) > 0 f( x,y) has a local minimum at ( p,q) D( p,q) > 0, fxx( p,q) < 0 f( x,y) has a local maximum at ( p,q) D( p,q) < 0, f( x,y) has a saddle  at ( p,q)

EXAMPLE 2    Identify the extrema and saddle points of f(x,y) = x²-y².      

Solution: Since f_x = 2x and f_y = -2y, the only critical point is ( 0,0) . However, f_xx = 2, f_yy = -2, and f_xy = 0, so that the discriminant of f is

D = fxxfyy-( fxy) 2 = ( 2) ( -2)-02 = -4 < 0

Thus, f( x,y) = x²-y² has a saddle at ( 0,0) .

EXAMPLE 3    Find the extrema and saddle points of f(x,y) = x³-3xy+y³.   

Solution: In example 1, we showed that the critical points of f are ( 0,0) and ( 1,1) . Since f_x(x,y) = 3x²-3y and f_y( x,y) = -3x+3y², the second derivatives of f( x,y) are

fxx = 6x,        fxy = -3,        fyy = 6y

Thus, the discriminant is

D( x,y) = ( 6x) ( 6y)  - ( -3)2 = 36xy-9

At ( 0,0) , we have D( 0,0) = 0 - 9 = -9 < 0. Thus, f has a saddle at ( 0,0) . At ( 1,1) , we have

D( 1,1) = 36·1·1 - 9 = 27 > 0

However, f_xx( 1,1) = 6 > 0, so f has a local minimum at ( 1,1) .

EXAMPLE 4    Find the local extrema and saddle points of

f( x,y) = xsin(xy)

Solution: The first partial derivatives are

fx = sin(xy) + xycos(xy),   fy =  x2cos( xy)

Setting f_y = 0 yields either x = 0 or cos( xy) = 0, the latter of which implies that

xy  =    p 2   +  np

for any integer n.  At such points, we would have f_x(x,y) either as 1 or -1 (but not 0). However, if y = 0, then

fx( x,0) = 0 + y

which implies that both f_x = 0 and f_y = 0 at (0, 0) (and nowhere else).  The second derivatives are

fxx = 2ycos(xy) - xy2sin(xy) fxy = fyx = 2xcos(xy) - x2ysin(xy) fyy = -x3 sin(xy)

and f_xx( 0,0) = f_xy( 0,0) = f_yy(0,0) = 0.  Thus, the discriminant is D( 0,0) = 0, so the second derivative test provides no information about the extrema or saddles of f( x,y) = xsin( xy) at (0,0) .

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Although it appears that there is a saddle at (0,0) in example 4, there is no way of determining this using the second derivative test.  Indeed, g(x,y) = x⁴ + y⁴ is positive everywhere except for g(0,0) = 0, so clearly g(x,y) has a minimum at (0,0).  But g_xx( 0,0) = g_xy( 0,0) = g_yy(0,0) = 0 implies that D(0,0) = 0, so this minimum cannot be identified using the second derivative test.