Question

Find the critical points and test for relative extrema. List the critical points for which the Second Partials Test fails. $$ f(x, y)=(x-1)^{2}(y+4)^{2} $$

Short answer

The critical point of the function is at (1, -4). However, the second partials test fails at this point, and the function's behavior cannot be classified based solely on this test.

Step-by-step solution

Compute the first partial derivatives

The first partial derivatives are found by differentiating the function with respect to each variable while keeping the other variable constant. Let's find \(f_{x}\) and \(f_{y}\):\n\(f_{x} = 2(x-1)(y+4)^{2}\)\n\(f_{y} = 2(x-1)^{2}(y+4)\)

Compute the critical points

Critical points occur where both partial derivatives are zero or undefined. So set \(f_{x}\) and \(f_{y}\) to zero and solve for x and y.\nFrom \(f_{x}=0\), we get \(x=1\) (since \(y+4\) cannot equal 0)\nFrom \(f_{y}=0\), we get \(y=-4\) (once more, \(x-1\) cannot equal 0)\nHence, the critical point is at (1, -4)

Compute the second partial derivatives

Second partial derivatives are computed by differentiating the first derivatives. Thus, we get:\n\(f_{xx} = 2(y+4)^{2}\)\n\(f_{xy} = 4(x-1)(y+4)\)\n\(f_{yy} = 2(x-1)^{2}\)

Perform the second partials test

Using the second partials test to classify the critical point, we compute the determinant of the Hessian matrix, \(D\), at the point (1, -4):\n\(D = f_{xx}f_{yy} - (f_{xy})^{2}\)\nSubstitute \(x=1\) and \(y=-4\) into \(D\), we find:\n\(D = 0\)

Classify the critical point

Since the determinant, \(D\), is equal to zero, the second partials test fails. Thus, we cannot classify the critical point as either a relative maximum, minimum, or saddle point based only on this test.