Animated Solution for Mathematics - Functions: A discontinuous function $y=f(x)$ satisfying $x^2+y^2=4$ is given by $f(x)=.........$

Imagine you are standing on the Cartesian plane, looking at the equation $x^2+y^2=4$. You see a perfect, symmetric circle centered at the origin with a radius of $2$. It is a beautiful, closed loop.

However, in the world of functions, this circle is a rebel. It fails the Vertical Line Test miserably. If you draw a vertical line anywhere between $x=-2$ and $x=2$, it pierces the circle at two distinct points.

For a single input $x$, you have two possible outputs for $y$. This is the definition of a relation, not a function. To tame this circle into a function, we must perform a surgical split.

We start by isolating $y$. From $x^2+y^2=4$, we get $y^2=4-x^2$. Taking the square root of both sides, we find:

This $\pm$ sign is our gateway to creativity. It splits the circle into two distinct branches: the upper semicircle, $y=\sqrt{4-x^2}$, and the lower semicircle, $y=-\sqrt{4-x^2}$.

Each of these, on its own, is a perfectly well-behaved, continuous function. But we want more than just a semicircle; we want to build a discontinuous function.

To create a jump discontinuity, we need to switch from one branch to the other abruptly. Let's choose the $y$-axis, where $x=0$, as our transition point.

We will define our function $f(x)$ to follow the upper branch for the left half of the circle and then, at the very moment we cross the $y$-axis, we will teleport to the lower branch. We define $f(x)$ as follows:

For the interval $[-2,0]$, we define $f(x)=\sqrt{4-x^2}$. At $x=0$, the value is $f(0)=\sqrt{4-0^2}=2$.

Now, for the interval $(0,2]$, we switch to the lower branch: $f(x)=-\sqrt{4-x^2}$. Notice the open interval at $x=0$; we must exclude it to ensure our function remains single-valued.

Now, let's verify our creation. Does it have a jump? We calculate the limits at the transition point $x=0$.

The left-hand limit is:

The right-hand limit is:

Since the left-hand limit of $2$ is not equal to the right-hand limit of $-2$, we have mathematically confirmed a jump discontinuity.

We have successfully taken a continuous relation and, through the power of piecewise definition, constructed a function that leaps across the $x$-axis. This is the elegance of calculus: taking the rigid geometry of a circle and bending it to our will to explore the nature of continuity itself.