Use the level curves in the figure to predict the location of the critical points of f and whether f has a saddle point or a local maximum or minimum at each critical point. Explain your reasoning. Then use the Second Derivatives Test to confirm your predictions.

4. f(x, y) = 3x − x³ − 2y² +3 y⁴

To predict: The location of the critical points of f and whether f has a local maximum, local minimum and saddle point at each critical point using the level curves in the figure. Use second derivative test to confirm the predictions.

Result used:

Second Derivative Test:

“Suppose the second partial derivatives of f are continuous on a disk with center (a,b) , and suppose that fx(a,b)=0 and fy(a,b)=0 (that is (a,b) is a critical point of f).

Let D=D(a,b)=fxx(a,b)fyy(a,b)−[fxy(a,b)]2

(a) If D>0 and fxx(a,b)>0 , then f(a,b) is a local minimum.

(b) If D>0 and fxx(a,b)<0 , then f(a,b) is a local maximum.

(c) If D<0 , then f(a,b) is not a local maximum or minimum and it is called a saddle point”.

Calculation:

From the given figure it is clear that the points (−1,1),(−1,−1) are surrounded by an oval shape.

Also, it can be identified that the value of f is increasing around all the points of (−1,±1) .

Thus, the local minimum occurs at (−1,±1) or near the point (−1,±1) . Therefore, the critical point is (−1,±1) .

The function is decreasing at the point (1,0) and hence it is local maximum

Moreover observe that near the points (−1,0),(1,1),(1,−1) , the values of f is increasing in some directions and decreasing in some other direction. So, there exists the saddle points at (−1,0),(1,1),(1,−1) .

Verification:

The given function is, f(x,y)=3x−x3−2y2+y4 .

Take the partial derivative in the given function with respect to x and obtain fx .

∂f∂x=∂∂x(3x−x3−2y2+y4)=∂∂x(3x)−∂∂x(x3)−∂∂x(2y2)+∂∂x(y4)=3(1)−3x2−0+0=3−3x2

Thus, ∂f∂x=3−3x2 . (1)

Take the partial derivative in the given function with respect to y and obtain fy .

∂f∂y=∂∂y(3x−x3−2y2+y4)=∂∂y(3x)−∂∂y(x3)−∂∂y(2y2)+∂∂y(y4)=0−0−2(2y)+4y3=−4y+4y3

Thus, ∂f∂y=−4y+4y3 . (2)

Set the above partial derivatives to 0 and find the values of x and y.

From the equation (1),

3−3x2=03x2=3x2=1x=±1

From the equation (2),

−4y+4y3=04y(y2−1)=0y=0,y2=1y=0,y=±1

Thus, the critical points are, (1,0),(1,1),(1,−1),(−1,0),(−1,1) and (−1,−1) .

Take the partial derivative of the equation (1) with respect to x and obtain fxx .

∂2f∂x2=∂∂x(3−3x2)=∂∂x(3)−∂∂x(3x2)=0−3(2x)=−6x

Hence, ∂2f∂x2=−6x .

Take the partial derivative of the equation (2) with respect to y and obtain fyy

∂2f∂y2=∂∂y(−4y+4y3)=∂∂y(−4y)+∂∂y(4y3)=−4(1)+4(3y2)=−4+12y2

Hence, ∂2f∂y2=−4+12y2 .

Take the partial derivative of the equation (1) with respect to y and obtain fxy .

∂2f∂x∂y=∂∂y(3−3x2)=∂∂y(3)−∂∂y(3x2)=0−0=0

Hence, ∂2f∂x∂y=0