Critical Point(s) of a Multivariable Function

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The discussion revolves around finding the critical points of the function f(x,y)=2+√(3(x-1)²+4(y+1)²). The partial derivatives are calculated as f_x(x,y)=3(x-1)/√(3(x-1)²+4(y+1)²) and f_y(x,y)=4(y+1)/√(3(x-1)²+4(y+1)²). Setting these derivatives to zero yields x=1 and y=-1, but these points do not exist within the function's domain. There is confusion regarding the definition of critical points, especially since Wolfram Alpha indicates that the function has no critical points. The key takeaway is that critical points must be within the function's domain, and in this case, (1,-1) is not valid.
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Homework Statement


Find the critical points of f.
f(x,y)=2+\sqrt{3(x-1)^2+4(y+1)^2}

Homework Equations


For fx(x,y) I get:
f_x(x,y)=0+\frac{1*6(x-1)}{2\sqrt{3(x-1)^2+4(y+1)^2}}=\frac{3(x-1)}{\sqrt{3(x-1)^2+4(y+1)^2}}
For fy(x,y) I get:
f_y(x,y)=0+\frac{1*8(y+1)}{2\sqrt{3(x-1)^2+4(y+1)^2}}=\frac{4(y+1)}{\sqrt{3(x-1)^2+4(y+1)^2}}

The Attempt at a Solution


Solving both of these for x and y when set equal to 0, gets me x = 1 and y = -1. However neither of these functions exist when x and y equal those values. Does the original function still have a critical point at (x,y) = (1,-1)?

Additionally, when I put the function into Wolfram Alpha it says that it has no critical points.
 
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What is the definition of the critical point?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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