fxx(5,3)fyy(5,3)= (-2)(-4), not (0)(0)!camino said:Homework Statement
Find all critical points of f(x,y) = 12xy-x^2 y-2xy^2 and test them for relative extrema using the 2nd partials test.
Homework Equations
The Attempt at a Solution
{0=fx(x,y)=12-2x-2 10x-2 x=5
{0=fy(x,y)=12-4y y=3
Critical point: (5,3)
fxx(x,y)= -2
fyy(x,y)= -4
fxy(x,y)= 0
d(5,3)=fxx(5,3)fyy(5,3)-(fxy(5,3))^2
d=(0)(0)-(0)
d=0 no conclusion
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Just wondering if I did this right?
Critical points are points on the graph of a mathematical function where the derivative of the function is equal to 0 or undefined. These points are important in determining the relative maximum and minimum values of the function.
The 2nd partials test is a mathematical tool used to determine if a critical point is a relative maximum, minimum, or saddle point. It involves finding the second derivative of the function at the critical point and evaluating it to see if it is positive, negative, or 0.
Yes, it is possible to have multiple critical points on a function. In fact, most functions have multiple critical points. These points can be found by setting the first derivative of the function equal to 0 and solving for the variable.
No, a critical point can only be one of these three types. The 2nd partials test helps determine which type of critical point it is by analyzing the concavity of the function at that point.
Critical points are important because they provide valuable information about the behavior of a function. They help us identify important points such as maximum and minimum values, which can be used in optimization problems. They also help us understand the shape and behavior of a function in different regions.