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Ind the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x,y)=2x

3 −6x+6xy 2
local maximum value(s) =
local minimum value(s) =
saddle point (s) =
(x,y,f)=________

1 Answer

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Final answer:

To find the local maximum and minimum values and saddle points of the function f(x,y) = 2x^3 - 6x + 6xy^2, we need to find the critical points and use the second partial derivatives test. The function has a local minimum at (1,0) and a saddle point at (-1,0).

Step-by-step explanation:

To find the local maximum and minimum values and saddle points of the function f(x,y) = 2x^3 - 6x + 6xy^2, we need to find the critical points where the first derivatives are equal to zero. Differentiating f(x,y) with respect to x and y, we get:

∂f/∂x = 6x^2 - 6 + 6y^2 = 0
∂f/∂y = 12xy = 0

Solving these equations simultaneously, we find the critical points (x, y) as (1, 0), (-1, 0), and (0, 0).

To determine whether each critical point is a local maximum, local minimum or saddle point, we need to use the second partial derivatives test. Evaluating the second derivatives at each critical point:

∂^2f/∂x^2 = 12x, ∂^2f/∂y^2 = 12x, ∂^2f/∂x∂y = 12y

Substituting the critical points (x, y) into these second derivatives, we find that:

For (1, 0): ∂^2f/∂x^2 = 12 > 0 (local minimum)
For (-1, 0): ∂^2f/∂x^2 = -12 < 0 (saddle point)
For (0, 0): ∂^2f/∂x^2 = 0 (test inconclusive)

Therefore, the local maximum value is DNE, the local minimum value is f(1,0)=2, and the saddle point is at (-1,0).

User Ashwin Yaprala
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