Saddle Points

Saddle points in a multivariable function are those critical points where the function attains neither a local maximum value nor a local minimum value. Saddle points mostly occur in multivariable functions. A few single variable functions like f(x) = x³ show a saddle point in its domain.

Critical points of a function are the points in the domain of the function where either the first derivative of the function is equal to zero or the derivative does not exist. If x = c is a critical point for a function f and f(c) exists, then c is called the critical point of f if either f’(c) = 0 or f’(c) does not exist.

Definition of Saddle Points

Saddle points of a multivariable function are those points in its domain where the tangent is parallel to the horizontal axis, but this point tends to be neither a local maximum nor a local minimum.

For a two-variable function f(x, y), its saddle point is defined as

Second Derivative Test for Saddle Points

Many times saddle points are not identifiable with the general second derivative test which is used to find the concavity or convexity of a function at any point. To find the saddle points within the domain of a function we require mixed partial derivatives.

Method of Finding of Saddle Points

Let f(x, y) be a function of two variables whose first and second-order partial derivatives exist and are continuous on a domain containing the point (a, b). Using the following steps apply the second derivative test to find the extreme points and saddle points.

• Determine the critical points (a, b) such that the partial derivatives f_x(a, b) = f_y(a, b) = 0. Discard those points for which any of the partial derivatives does not exist.
• Calculate the discriminant D = f_xx(a, b) f_yy (a, b) – [f_xy(a, b)]² for each critical point.
• Check for the cases:
  • If D > 0 and f_xx(a, b) > 0, f has a local minimum at (a, b)
  • If D > 0 and f_xx(a, b) < 0, has a local maximum at (a, b)
  • If D < 0, f has a saddle point (a, b)
  • If D = 0, the test fails.

Related Articles

• Continuity of Functions
• Maxima and Minima
• Inflection point
• Second Derivative Test
• Inverse Functions

Solved Examples on Saddle Points

Example 1:

Find the critical points of the function f(x, y) = y² – x² and check for the presence of any saddle point.

Solution:

Let us find the first-order partial derivative with respect to x and y to determine the critical points.

f_x(x, y) = –2x

and f_y(x, y) = 2y which exists everywhere.

Clearly, f has a critical point at (0, 0). Let us apply the second-order partial derivative test,

f_xx(x, y) = –2

f_yy(x, y) = 2

f_xy(x, y) = 0

Therefore, the discriminant D = f_xx(0, 0) f_yy (0, 0) – [f_xy(0, 0)]²

⇒ D = – 4 – 0² = – 4 < 0

Hence, at (0, 0) f has a saddle point.

Example 2:

Find the critical points of the function f(x, y) = x²y² – 5x² – 5y² – 8xy and check for the presence of any saddle point.

Solution:

Let us find the first-order partial derivative with respect to x and y to determine the critical points.

f_x(x, y) = 2xy² – 10x – 8y

and f_y(x, y) = 2x²y – 10y – 8x which exists everywhere.

Now, the second equation from first and equating to zero, we get

2xy (y – x) + 2(y – x) = 0

⇒ (y – x)(xy + 1) = 0

⇒ either x = y or y = – 1/x

If x = y, substituting this in any of the partial derivatives, we get

2x³ – 18x = 0

⇒ x = 0, 3, –3

And if y = – 1/x, substituting this in any of the partial derivatives, we get

2/x – 10x + 8/x = 0 ⇒ 1 – x² = 0

⇒ x = 1, –1

Therefore, we get the critical points (0, 0), (3, 3), ( – 3, –3), (1, –1) and ( –1, 1).

Now, f_xx(x, y) = 2y² – 10

f_yy(x, y) = 2x² – 10

f_xy(x, y) = 4xy – 8

At (0, 0), D = f_xx(0, 0) f_yy (0, 0) – [f_xy(0, 0)]² = 100 – 64 = 36 > 0 and f_xx(0, 0) = – 10 < 0

Hence, at (0, 0) we have a local maximum

At (±3, ±3), D = f_xx(±3, ±3) f_yy (±3, ±3) – [f_xy(±3, ±3)]² = 64 – 784 = – 720 < 0

Hence, at (±3, ±3) we have saddle points.

At (±1, ±1), D = f_xx(±1, ±1) f_yy (±1, ±1) – [f_xy(±1, ±1)]² = 64 – 144 = a negative quantity

Hence, at (±1, ±1) also we have saddle points.