1)Find 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) = y^2 − 10y cos x,&nbsp;&nbsp;&nbsp; −1 ≤ x ≤ 72)Find 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) = 7(x^2 + y^2)e^(y2 − x2)

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1)Find 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) = y^2 − 10y cos x,&amp;nbsp;&amp;nbsp;&amp;nbsp; −1 ≤ x ≤ 72)Find 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) = 7(x^2 + y^2)e^(y2 − x2)

madeleinejames20 · Accepted Answer

To find the local maximum and minimum values and saddle points of the function f(x,y)=y2−10ycos(x), we need to compute its first and second partial derivatives and analyze the critical points.Let&#039;s begin by finding the first partial derivatives:∂f∂x=10ysin(x)∂f∂y=2y−10cos(x)To find the critical points, we set both partial derivatives equal to zero and solve the resulting system of equations:10ysin(x)=02y−10cos(x)=0From the first equation, we have two possibilities:1. 10y=0⟹y=02. sin(x)=0⟹x=kπ, where k is an integerSubstituting these possibilities into the second equation:For y=0, we get −10cos(x)=0⟹cos(x)=0⟹x=π2+kπ, where k is an integer.For sin(x)=0, we get 2y−10=0⟹y=102=5.Therefore, the critical points are (π2+kπ,0) and (kπ,5), where k is an integer.To determine the nature of these critical points, we need to compute the second partial derivatives:∂2f∂x2=10ycos(x)∂2f∂y2=2∂2f∂x∂y=0Now, we can evaluate the second partial derivatives at the critical points.For the critical point (π2+kπ,0):∂2f∂x2=0∂2f∂y2=2∂2f∂x∂y=0For the critical point (kπ,5):∂2f∂x2=10(5)cos(kπ)=(−1)k·50∂2f∂y2=2∂2f∂x∂y=0To classify the critical points, we can use the second partial derivative test:1. For a critical point (π2+kπ,0), the second partial derivative test is inconclusive.2. For a critical point (kπ,5):- When k is even, the second partial derivative ∂2f∂x2=50&amp;gt;0, so we have a local minimum.- When k is odd, the second partial derivative ∂2f∂x2=−50&amp;lt;0, so we have a local maximum.Therefore, the local maximum and minimum values of the function f(x,y) are obtained at the critical points (kπ,5), where k is an odd integer.As for the saddle points, since the second partial derivative test is inconclusive for the critical points (π2+kπ,0), we cannot determine the presence of saddle points based on this analysis.To graph the function f(x,y)=y2−10ycos(x), it is recommended to use three-dimensional graphing software to visualize the function&#039;s behavior accurately.

1)Find the local maximum and minimum values and

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