Is Abs Xy Abs X Abs Y Continuous at the Origin

Let's first get a quick picture of the rectangle for reference purposes.

The boundary of this rectangle is given by the following conditions.

\[\begin{align*}&{\mbox{right side :}}\hspace{0.25in}x = 1,\,\, - 1 \le y \le 1\\ & {\mbox{left side :}}\hspace{0.25in}x = - 1,\,\, - 1 \le y \le 1\\ & {\mbox{upper side :}}\hspace{0.25in}y = 1,\,\, - 1 \le x \le 1\\ & {\mbox{lower side :}}\hspace{0.25in}y = - 1,\,\, - 1 \le x \le 1\end{align*}\]

These will be important in the second step of our process.

We'll start this off by finding all the critical points that lie inside the given rectangle. To do this we'll need the two first order derivatives.

\[{f_x} = 2x - 4xy\hspace{0.5in}{f_y} = 8y - 2{x^2}\]

Note that since we aren't going to be classifying the critical points we don't need the second order derivatives. To find the critical points we will need to solve the system,

\[\begin{align*}2x - 4xy & = 0\\ 8y - 2{x^2} & = 0\end{align*}\]

We can solve the second equation for \(y\) to get,

\[y = \frac{{{x^2}}}{4}\]

Plugging this into the first equation gives us,

\[2x - 4x\left( {\frac{{{x^2}}}{4}} \right) = 2x - {x^3} = x\left( {2 - {x^2}} \right) = 0\]

This tells us that we must have \(x = 0\) or \(x = \pm \sqrt 2 = \pm 1.414...\). Now, recall that we only want critical points in the region that we're given. That means that we only want critical points for which \( - 1 \le x \le 1\). The only value of \(x\) that will satisfy this is the first one so we can ignore the last two for this problem. Note however that a simple change to the boundary would include these two so don't forget to always check if the critical points are in the region (or on the boundary since that can also happen).

Plugging \(x = 0\) into the equation for \(y\) gives us,

\[y = \frac{{{0^2}}}{4} = 0\]

The single critical point, in the region (and again, that's important), is \(\left( {0,0} \right)\). We now need to get the value of the function at the critical point.

\[f\left( {0,0} \right) = 4\]

Eventually we will compare this to values of the function found in the next step and take the largest and smallest as the absolute extrema of the function in the rectangle.

Now we have reached the long part of this problem. We need to find the absolute extrema of the function along the boundary of the rectangle. What this means is that we're going to need to look at what the function is doing along each of the sides of the rectangle listed above.

Let's first take a look at the right side. As noted above the right side is defined by

\[x = 1,\,\, - 1 \le y \le 1\]

Notice that along the right side we know that \(x = 1\). Let's take advantage of this by defining a new function as follows,

\[g\left( y \right) = f\left( {1,y} \right) = {1^2} + 4{y^2} - 2\left( {{1^2}} \right)y + 4 = 5 + 4{y^2} - 2y\]

Now, finding the absolute extrema of \(f\left( {x,y} \right)\) along the right side will be equivalent to finding the absolute extrema of \(g\left( y \right)\) in the range \( - 1 \le y \le 1\). Hopefully you recall how to do this from Calculus I. We find the critical points of \(g\left( y \right)\) in the range \( - 1 \le y \le 1\) and then evaluate \(g\left( y \right)\) at the critical points and the end points of the range of \(y\)'s.

Let's do that for this problem.

\[g'\left( y \right) = 8y - 2\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\,y = \frac{1}{4}\]

This is in the range and so we will need the following function evaluations.

\[g\left( { - 1} \right) = 11\hspace{0.5in}g\left( 1 \right) = 7\hspace{0.5in}g\left( {\frac{1}{4}} \right) = \frac{{19}}{4} = 4.75\]

Notice that, using the definition of \(g\left( y \right)\) these are also function values for \(f\left( {x,y} \right)\).

\[\begin{align*}g\left( { - 1} \right) & = f\left( {1, - 1} \right) = 11\hspace{0.5in}\\ g\left( 1 \right) & = f\left( {1,1} \right) = 7\\ g\left( {\frac{1}{4}} \right) & = f\left( {1,\frac{1}{4}} \right) = \frac{{19}}{4} = 4.75\end{align*}\]

We can now do the left side of the rectangle which is defined by,

\[x = - 1,\,\, - 1 \le y \le 1\]

Again, we'll define a new function as follows,

\[g\left( y \right) = f\left( { - 1,y} \right) = {\left( { - 1} \right)^2} + 4{y^2} - 2{\left( { - 1} \right)^2}y + 4 = 5 + 4{y^2} - 2y\]

Notice however that, for this boundary, this is the same function as we looked at for the right side. This will not always happen, but since it has let's take advantage of the fact that we've already done the work for this function. We know that the critical point is \(y = \frac{1}{4}\) and we know that the function value at the critical point and the end points are,

\[g\left( { - 1} \right) = 11\hspace{0.5in}g\left( 1 \right) = 7\hspace{0.5in}g\left( {\frac{1}{4}} \right) = \frac{{19}}{4} = 4.75\]

The only real difference here is that these will correspond to values of \(f\left( {x,y} \right)\) at different points than for the right side. In this case these will correspond to the following function values for \(f\left( {x,y} \right)\).

\[\begin{align*}g\left( { - 1} \right) & = f\left( { - 1, - 1} \right) = 11\hspace{0.5in}\\ g\left( 1 \right) & = f\left( { - 1,1} \right) = 7\\ g\left( {\frac{1}{4}} \right) & = f\left( { - 1,\frac{1}{4}} \right) = \frac{{19}}{4} = 4.75\end{align*}\]

We can now look at the upper side defined by,

\[y = 1,\,\, - 1 \le x \le 1\]

We'll again define a new function except this time it will be a function of \(x\).

\[h\left( x \right) = f\left( {x,1} \right) = {x^2} + 4\left( {{1^2}} \right) - 2{x^2}\left( 1 \right) + 4 = 8 - {x^2}\]

We need to find the absolute extrema of \(h\left( x \right)\) on the range \( - 1 \le x \le 1\). First find the critical point(s).

\[h'\left( x \right) = - 2x\hspace{0.5in} \Rightarrow \hspace{0.5in}x = 0\]

The value of this function at the critical point and the end points is,

\[h\left( { - 1} \right) = 7\hspace{0.5in}h\left( 1 \right) = 7\hspace{0.5in}h\left( 0 \right) = 8\]

and these in turn correspond to the following function values for \(f\left( {x,y} \right)\)

\[\begin{align*}h\left( { - 1} \right) & = f\left( { - 1,1} \right) = 7\\ h\left( 1 \right) & = f\left( {1,1} \right) = 7\\ h\left( 0 \right) & = f\left( {0,1} \right) = 8\end{align*}\]

Note that there are several "repeats" here. The first two function values have already been computed when we looked at the right and left side. This will often happen.

Finally, we need to take care of the lower side. This side is defined by,

\[y = - 1,\,\, - 1 \le x \le 1\]

The new function we'll define in this case is,

\[h\left( x \right) = f\left( {x, - 1} \right) = {x^2} + 4{\left( { - 1} \right)^2} - 2{x^2}\left( { - 1} \right) + 4 = 8 + 3{x^2}\]

The critical point for this function is,

\[h'\left( x \right) = 6x\hspace{0.5in} \Rightarrow \hspace{0.5in}x = 0\]

The function values at the critical point and the endpoint are,

\[h\left( { - 1} \right) = 11\hspace{0.5in}h\left( 1 \right) = 11\hspace{0.5in}h\left( 0 \right) = 8\]

and the corresponding values for \(f\left( {x,y} \right)\) are,

\[\begin{align*}h\left( { - 1} \right) & = f\left( { - 1, - 1} \right) = 11\\ h\left( 1 \right) & = f\left( {1, - 1} \right) = 11\\ h\left( 0 \right) & = f\left( {0, - 1} \right) = 8\end{align*}\]

The final step to this (long…) process is to collect up all the function values for \(f\left( {x,y} \right)\) that we've computed in this problem. Here they are,

\[\begin{align*}f\left( {0,0} \right) & = 4\hspace{0.5in} & f\left( {1, - 1} \right) & = 11\hspace{0.5in} & f\left( {1,1} \right) & = 7\\ f\left( {1,\frac{1}{4}} \right) & = 4.75\hspace{0.5in} & f\left( { - 1,1} \right) & = 7\hspace{0.5in} & f\left( { - 1, - 1} \right) & = 11\\ f\left( { - 1,\frac{1}{4}} \right) & = 4.75\hspace{0.25in} & f\left( {0,1} \right) & = 8\hspace{0.5in} & f\left( {0, - 1} \right) & = 8\end{align*}\]

The absolute minimum is at \(\left( {0,0} \right)\) since gives the smallest function value and the absolute maximum occurs at \(\left( {1, - 1} \right)\) and \(\left( { - 1, - 1} \right)\) since these two points give the largest function value.

Here is a sketch of the function on the rectangle for reference purposes.

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Source: https://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx

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