Examples from Local extrema All examples from Local extrema.
[ EXAMPLE ] Let
f ( x , y ) = 8 x 3 + y 2 − 6 x y − 2 y + 3. f(x,y)=8x^3+y^2-6xy-2y+3. f ( x , y ) = 8 x 3 + y 2 − 6 x y − 2 y + 3. Then
f x = 24 x 2 − 6 y , f y = 2 y − 6 x − 2. f_x=24x^2-6y,
\qquad
f_y=2y-6x-2. f x = 24 x 2 − 6 y , f y = 2 y − 6 x − 2. The stationary equations give
y = 4 x 2 , y = 3 x + 1. y=4x^2,
\qquad
y=3x+1. y = 4 x 2 , y = 3 x + 1. Hence
4 x 2 = 3 x + 1 , ( 4 x + 1 ) ( x − 1 ) = 0. 4x^2=3x+1,
\qquad
(4x+1)(x-1)=0. 4 x 2 = 3 x + 1 , ( 4 x + 1 ) ( x − 1 ) = 0. The stationary points are
( 1 , 4 ) and ( − 1 4 , 1 4 ) . {(1,4)}
\qquad\text{and}\qquad
{\left(-\frac14,\frac14\right)}. ( 1 , 4 ) and ( − 4 1 , 4 1 ) . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] Define
f ( x , y ) = { x y x 2 + y 2 , ( x , y ) ≠ ( 0 , 0 ) , 0 , ( x , y ) = ( 0 , 0 ) . f(x,y)=
\begin{cases}
\dfrac{xy}{x^2+y^2},&(x,y)\ne(0,0),\\[6pt]
0,&(x,y)=(0,0).
\end{cases} f ( x , y ) = ⎩ ⎨ ⎧ x 2 + y 2 x y , 0 , ( x , y ) = ( 0 , 0 ) , ( x , y ) = ( 0 , 0 ) . Along the coordinate axes,
f ( h , 0 ) = 0 , f ( 0 , h ) = 0 , f(h,0)=0,
\qquad
f(0,h)=0, f ( h , 0 ) = 0 , f ( 0 , h ) = 0 , so
f x ( 0 , 0 ) = 0 , f y ( 0 , 0 ) = 0. f_x(0,0)=0,
\qquad
f_y(0,0)=0. f x ( 0 , 0 ) = 0 , f y ( 0 , 0 ) = 0. However, along y = x y=x y = x
f ( t , t ) = t 2 2 t 2 = 1 2 ( t ≠ 0 ) . f(t,t)=\frac{t^2}{2t^2}=\frac12
\qquad(t\ne0). f ( t , t ) = 2 t 2 t 2 = 2 1 ( t = 0 ) . The function is therefore not continuous at the origin and cannot be
differentiable there, even though its two partial derivatives exist and
vanish.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
f ( x , y ) = x y , f(x,y)=xy, f ( x , y ) = x y , the only stationary point is ( 0 , 0 ) (0,0) ( 0 , 0 )
∇ f ( x , y ) = ( y x ) . \nabla f(x,y)
=
\begin{pmatrix}
y\\
x
\end{pmatrix}. ∇ f ( x , y ) = ( y x ) . Along y = x y=x y = x
f ( t , t ) = t 2 > 0 ( t ≠ 0 ) , f(t,t)=t^2>0
\qquad(t\ne0), f ( t , t ) = t 2 > 0 ( t = 0 ) , whereas along y = − x y=-x y = − x
f ( t , − t ) = − t 2 < 0. f(t,-t)=-t^2<0. f ( t , − t ) = − t 2 < 0. Every neighbourhood of the origin contains larger and smaller values than
f ( 0 , 0 ) = 0 f(0,0)=0 f ( 0 , 0 ) = 0
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
f ( x , y ) = x 2 + y 2 , f(x,y)=x^2+y^2, f ( x , y ) = x 2 + y 2 , the Hessian at the origin is
H f ( 0 , 0 ) = ( 2 0 0 2 ) , D = 4 > 0 , f x x = 2 > 0. H_f(0,0)
=
\begin{pmatrix}
2&0\\
0&2
\end{pmatrix},
\qquad
D=4>0,
\qquad
f_{xx}=2>0. H f ( 0 , 0 ) = ( 2 0 0 2 ) , D = 4 > 0 , f xx = 2 > 0. Thus the origin is a local minimum; f ( x , y ) ≥ 0 f(x,y)\ge0 f ( x , y ) ≥ 0
For
g ( x , y ) = − x 2 − y 2 , g(x,y)=-x^2-y^2, g ( x , y ) = − x 2 − y 2 , we have D = 4 > 0 D=4>0 D = 4 > 0 g x x = − 2 < 0 g_{xx}=-2<0 g xx = − 2 < 0
For
h ( x , y ) = x 2 − y 2 , h(x,y)=x^2-y^2, h ( x , y ) = x 2 − y 2 , the Hessian is
H h ( 0 , 0 ) = ( 2 0 0 − 2 ) , D = − 4 < 0. H_h(0,0)
=
\begin{pmatrix}
2&0\\
0&-2
\end{pmatrix},
\qquad
D=-4<0. H h ( 0 , 0 ) = ( 2 0 0 − 2 ) , D = − 4 < 0. The origin is therefore a saddle; the two coordinate axes show this directly.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let
f ( x , y ) = x 3 − y 3 + 3 x y . f(x,y)=x^3-y^3+3xy. f ( x , y ) = x 3 − y 3 + 3 x y . The first derivatives are
f x = 3 x 2 + 3 y , f y = − 3 y 2 + 3 x . f_x=3x^2+3y,
\qquad
f_y=-3y^2+3x. f x = 3 x 2 + 3 y , f y = − 3 y 2 + 3 x . Thus
y = − x 2 , x = y 2 , y=-x^2,
\qquad
x=y^2, y = − x 2 , x = y 2 , so x = x 4 x=x^4 x = x 4 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , − 1 ) (1,-1) ( 1 , − 1 )
The Hessian is
H f ( x , y ) = ( 6 x 3 3 − 6 y ) . H_f(x,y)
=
\begin{pmatrix}
6x&3\\
3&-6y
\end{pmatrix}. H f ( x , y ) = ( 6 x 3 3 − 6 y ) . At ( 0 , 0 ) (0,0) ( 0 , 0 )
det H f ( 0 , 0 ) = − 9 < 0 , \det H_f(0,0)=-9<0, det H f ( 0 , 0 ) = − 9 < 0 , so the origin is a saddle. At ( 1 , − 1 ) (1,-1) ( 1 , − 1 )
H f ( 1 , − 1 ) = ( 6 3 3 6 ) , det H f ( 1 , − 1 ) = 27 > 0 , f x x ( 1 , − 1 ) = 6 > 0. H_f(1,-1)
=
\begin{pmatrix}
6&3\\
3&6
\end{pmatrix},
\qquad
\det H_f(1,-1)=27>0,
\qquad
f_{xx}(1,-1)=6>0. H f ( 1 , − 1 ) = ( 6 3 3 6 ) , det H f ( 1 , − 1 ) = 27 > 0 , f xx ( 1 , − 1 ) = 6 > 0. Therefore ( 1 , − 1 ) (1,-1) ( 1 , − 1 )
f ( 1 , − 1 ) = 1 − ( − 1 ) 3 + 3 ( 1 ) ( − 1 ) = − 1. f(1,-1)=1-(-1)^3+3(1)(-1)=-1. f ( 1 , − 1 ) = 1 − ( − 1 ) 3 + 3 ( 1 ) ( − 1 ) = − 1. [RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let
f ( x , y ) = x y e x 2 + y 2 . f(x,y)=xy\,e^{x^2+y^2}. f ( x , y ) = x y e x 2 + y 2 . The derivatives factor as
f x = y e x 2 + y 2 ( 1 + 2 x 2 ) , f y = x e x 2 + y 2 ( 1 + 2 y 2 ) . f_x=y\,e^{x^2+y^2}(1+2x^2),
\qquad
f_y=x\,e^{x^2+y^2}(1+2y^2). f x = y e x 2 + y 2 ( 1 + 2 x 2 ) , f y = x e x 2 + y 2 ( 1 + 2 y 2 ) . The exponential factors and the terms 1 + 2 x 2 1+2x^2 1 + 2 x 2 1 + 2 y 2 1+2y^2 1 + 2 y 2 ( 0 , 0 ) (0,0) ( 0 , 0 )
H f ( 0 , 0 ) = ( 0 1 1 0 ) , det H f ( 0 , 0 ) = − 1. H_f(0,0)
=
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix},
\qquad
\det H_f(0,0)=-1. H f ( 0 , 0 ) = ( 0 1 1 0 ) , det H f ( 0 , 0 ) = − 1. The origin is a saddle. In this example, the derivative factors prove that
the positive exponential adds no stationary points; positivity of a factor
alone would not guarantee that for every product.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
f ( x , y ) = 8 x 3 + y 2 − 6 x y − 2 y + 3 , f(x,y)=8x^3+y^2-6xy-2y+3, f ( x , y ) = 8 x 3 + y 2 − 6 x y − 2 y + 3 , the Hessian and its determinant are
H f ( x , y ) = ( 48 x − 6 − 6 2 ) , det H f ( x , y ) = 96 x − 36. H_f(x,y)
=
\begin{pmatrix}
48x&-6\\
-6&2
\end{pmatrix},
\qquad
\det H_f(x,y)=96x-36. H f ( x , y ) = ( 48 x − 6 − 6 2 ) , det H f ( x , y ) = 96 x − 36. At ( 1 , 4 ) (1,4) ( 1 , 4 )
det H f ( 1 , 4 ) = 60 > 0 , f x x ( 1 , 4 ) = 48 > 0 , \det H_f(1,4)=60>0,
\qquad
f_{xx}(1,4)=48>0, det H f ( 1 , 4 ) = 60 > 0 , f xx ( 1 , 4 ) = 48 > 0 , so ( 1 , 4 ) (1,4) ( 1 , 4 ) ( − 1 / 4 , 1 / 4 ) (-1/4,1/4) ( − 1/4 , 1/4 )
det H f ( − 1 4 , 1 4 ) = − 60 < 0 , \det H_f\left(-\frac14,\frac14\right)=-60<0, det H f ( − 4 1 , 4 1 ) = − 60 < 0 , so that point is a saddle.
[RIGHT CLICK] [ASK AI] Local extrema Builds on Multivariable approximation .