Question at position 1 The number of critical points of f(x,y)=x2+x2y+y2−2y+2f\left(x,y\right)=x^2+x^2y+y^2-2y+2 is42301

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Question at position 1 The number of critical points of f(x,y)=x2+x2y+y2−2y+2f\left(x,y\right)=x^2+x^2y+y^2-2y+2 is42301

BlackTom AI · Accepted Answer

We start by examining the function: f(x,y) = x^2 + x^2 y + y^2 - 2y + 2.
To find critical points, we compute the partial derivatives:
- f_x = ∂f/∂x = 2x + 2xy = 2x(1 + y).
- f_y = ∂f/∂y = x^2 + 2y - 2.
Critical points satisfy f_x = 0 and f_y = 0.
From f_x = 0, we have two cases: either x = 0 or y = -1.
Case A: x = 0. Plugging into f_y = 0 gives 0^2 + 2y - 2 = 0 ⇒ 2y = 2 ⇒ y = 1. So one critical point is (0, 1).
Case B: y = -1. Plugging into f_y = 0 gives x^2 + 2(-1) - 2 = 0 ⇒ x^2 - 4 = 0 ⇒ x = ±2. So two more critical points are (2, -1) and (-2, -1).
In total, there are 3 critical points: (0, 1), (2, -1), and (-2, -1).
Now evaluate each answer option:
- Option "4": This claims there are four critical points. This is incorrect because the system above yields only three solutions for (x, y).
- Option "2": This claims there are two critical points. This is incorrect for the same reason; we found three distinct critical points.
- Option "3": This claims there are three critical points. This matches the computed result from solving f_x = 0 and f_y = 0, since we obtained exactly three distinct solutions.
- Option "0": This claims there are no critical points. This is clearly false because the gradient vanishes at the points identified in Case A and Case B.
- Option "1": This claims there is exactly one critical point. This is also false given multiple solutions from the equations.
Thus, the only option aligned with the correct count is the one stating three critical points, and the reasoning above shows why the other counts are not supported by the gradient equations.

Question at position 1 The number of critical points of f(x,y)=x2+x2y+y2−2y+2f\left(x,y\right)=x^2+x^2y+y^2-2y+2 is42301

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