# What are the critical points, if any, of #f(x,y) = 2x^3 - 6x^2 + y^3 + 3y^2 - 48x - 45y#?

# {: (f(-2,-5),= 231,=>(-2,-5,231)), (f(-2,3),= -25,=>(-2,3,25)), (f(4,-5),= 15,=>(4,-5,15)), (f(4,3), = -241,=>(4,3,-241)) :} #

The theory to identify the extrema of

- Solve simultaneously the critical equations
# (partial f) / (partial x) =(partial f) / (partial y) =0 \ \ \ # (ie#f_x=f_y=0# ) - Evaluate
#f_(x x), f_(yy)# and#f_(xy) (=f_(yx))# at each of these critical points. Hence evaluate#Delta=f_(x x)f_(yy)-f_(xy)^2# at each of these points - Determine the nature of the extrema;
# {: (Delta>0, "There is minimum if " f_(x x)>0),(, "and a maximum if " f_(x x)<0), (Delta<0, "there is a saddle point"), (Delta=0, "Further analysis is necessary") :} #

So we have:

# f(x,y) = 2x^3 - 6x^2 + y^3 + 3y^2 - 48x - 45y # Let us find the first partial derivatives:

# (partial f) / (partial x) = 6x^2-12x-48 #

# (partial f) / (partial y) = 3y^2+6y-45 # So our critical simultaneous equations are:

# (partial f) / (partial x) = 0 => 6x^2-12x-48 = 0 #

# (partial f) / (partial y) = 0 => 3y^2+6y-45 = 0 #

# :. x^2 - 2x - 8 = 0 #

# :. (x-4)(x+2) = 0 #

# :. x = -2, 4 #

# :. y^2 + 2y - 15 = 0 #

# :. (y+5)(y-3) = 0 #

# :. y = -5, 3 # Any permutation of these solutions will simultaneously make both partial derivatives vanish, so the critical values are:

#(x,y)=(-2,-5), (-2,3), (4,-5), (4,3) # So we can now calculate the coordinates of the critical points:

# {: (f(-2,-5),= 231,=>(-2,-5,231)), (f(-2,3),= -25,=>(-2,3,25)), (f(4,-5),= 15,=>(4,-5,15)), (f(4,3), = -241,=>(4,3,-241)) :} # We can visualise these critical points on a 3D-plot:

As we were not asked to determine the nature of the critical points I will omit that analysis.

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To find the critical points of the function (f(x, y) = 2x^3 - 6x^2 + y^3 + 3y^2 - 48x - 45y), we need to find where the partial derivatives with respect to (x) and (y) are both zero.

Taking partial derivatives with respect to (x) and (y), and setting them equal to zero, we get:

[\frac{\partial f}{\partial x} = 6x^2 - 12x - 48 = 0] [\frac{\partial f}{\partial y} = 3y^2 + 6y - 45 = 0]

Solving these equations will give us the critical points of the function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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