520680…: “Over a teal parallelogram”

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Suppose that $F(x,y) = \begin {cases} 4x^2y , & x+y > 1 \\ 2x, & x+y \leq 1 \end {cases}$ .

Answer the following questions.

True or False:

$\Lim {x,y}{-1,1} F(x,y)= \Lim {x,y}{-1,1} 2x = -2$ . TrueFalse

Near the point $(-1,1)$ , the function $F(x,y)$ is defined by $F(x,y) = 2x$ for all $(x,y)$ .

The domain of the function, $\R ^2$ , is shown as well as the

formula used to find $F(x,y)$ for each $(x,y)$ in the domain.

More precisely, this means that we can squeeze an entire open ball centered at $(-1,1)$ in which $F(x,y) = 2x$ for any $(x,y)$ in the open ball. To construct the open ball, we just need to use a radius so the ball does not intersect the line $x+y=1$ . From the image provided, it is clear that we can find such a ball.

True or False:

$F(-1,2) = 2(-1)+2$ , so $\Lim {x,y}{-1,2} F(x,y)= \Lim {x,y}{-1,2} 2x = -2$ . TrueFalse

Near the point $(-1,2)$ , the function $F(x,y)$ is not defined by $F(x,y) = 2x$ for all $(x,y)$ ; at any point lower than and/or left of $(-1,2)$ , we have $F(x,y) = 2x$ , but for any point higher than and/or right of $(-1,2)$ , we have that $F(x,y) =4x^2y$ .

The domain of the function, $\R ^2$ , is shown as well as the

formula used to find $F(x,y)$ for each $(x,y)$ in the domain.

More precisely, this means that no open ball centered at $(-1,2)$ contains points for $F(x,y) = 2x$ for any $(x,y)$ in the open ball. From the image provided, it is clear that no matter how small the radius of a ball centered at $(-1,2)$ may be, it will always contain points in both the blue and red regions.

Compute $\Lim {x,y}{-1,2} F(x,y)$ or conclude that it does not exist by following the steps below.

If we approach $(-1,2)$ along the line $x=-1$ and take $y<2$ , $F(x,y) = \answer {2x}$ . Thus $F(x,y) \to \answer {-2}$ as $\vector {x,y} \to \vector {-1,2}$ along this path.
If we approach $(-1,2)$ along the line $x=-1$ and take $y>2$ , $F(x,y) = \answer {4x^2y}$ . Thus $F(x,y) \to \answer {8}$ as $\vector {x,y} \to \vector {-1,2}$ along this path.

Does $\Lim {x,y}{-1,2} F(x,y)$ exist? YesNo

Is the function continuous at any point $(x,y)$ for which $x+y \neq 1$ ? YesNo

Is the function discontinuous at any point for which $x+y =1$ ? YesNo

As long as $x+y \neq 1$ , the function is defined by the same formula for all $(x,y)$ in a suitable open ball centered at $(x,y)$ , so $\Lim {x,y}{a,b}F(x,y) = F(a,b)$ . Since each piece is continuous, this means $F(x,y)$ is continuous at all $(a,b)$ for which $a+b \neq 1$ .

Along the line $x+y=1$ , we have to check whether the pieces “line up” to give the same $z$ -value.

If $x+y=1$ , how many ordered pair $(x,y)$ are there for which $F(x,y)$ is continuous?

None One Two Three More than three, but finitely many Infinitely many

To find these points, we must have

$4x^2y=2x.$ and noting that along the line $x+y=1$ , we have $y=1-x$ and thus must find all $x$ so

$4x^2(1-x)=2x.$

Simplifying and rearranging this gives

$\answer {-4}x^3+\answer {4}x^2-\answer {2}x = 0.$ (don’t just cancel out the $x$ on each side; doing so eliminates one of the solutions!)

This means either $x=0$ or $\answer {-4x^2+4x}-2=0$ .

For $x=0$ , we find that $y=\answer {1}$ , so the function is continuous at $\left (0,\answer {1}\right )$ .
The quadratic equation can be used to show that there are real solutions to the quadratic equation $-4x^2+4x-2=0$ .