Jump Discontinuity

For the function in the graph below,

1. Find an equation for the function
2. Show that it has a jump discontinuity at $x=2$.

Function with a jump discontinuity, StudySmarter Original

Answer:

When writing the equation of a function like this, it is easier to think of it in parts and then put all the parts together at the end.

1. To the left of $x=2$, the function has a y-intercept at $\left(0,2\right)$ and the slope of the line is 1, so the equation of the line on the left in slope-intercept form is $y=x+\hspace{0.17em}2$.
2. To the right of $x=2$, the function has an x-intercept at $(3,0)$ and the slope of the line is $-1$. So the equation of the line on the right is $y=-x+3$.
3. At the point $x=2$ the function value is 1.
4. Putting all of those parts together, you get the piecewise defined function

$f\left(x\right)=\left\{\begin{array}{ll}x+\hspace{0.17em}2,& x<\hspace{0.17em}2\\ -x+3,& x\ge 2\end{array}\right.$.

To show that it has a jump discontinuity, unfortunately it isn't sufficient to point to the spot on the graph and say "see, it jumps!". Instead you need to look at the limits from the left and right. Using the function definition you found,

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\left(x+2\right)=2+2=4$,

and

$\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\left(-x+3\right)=-2+3=1$.

So you can see that both the limit from the left and the limit from the right are real numbers, but they certainly aren't the same number. That shows there is a jump discontinuity at $x=2$.

Does the function

$f\left(x\right)=\left\{\begin{array}{ll}{x}^2,& x<1\\ 0,& x=1\\ 2-(x^2-1),& x>\hspace{0.17em}1\end{array}\right.$

have a jump discontinuity at $x=1$?

Answer:

Let's check the limits:

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }{x}^2=1$

and

id="2752728" role="math" $\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }(2-\left(x^2-1\right)=2-\left(1^2-1\right)=2-0=2$,

The left-handed limit is not equal to the right-handed limit at $x=1$, so you know that not only is the function discontinuous at $x=1$, it has a jump discontinuity there.

Does the function

$g\left(x\right)=\left\{\begin{array}{ll}{x}^2,& x<1\\ 2x-1,& x\ge 1\end{array}\right.$

Answer:

The idea is again to check the limits:

$\underset{x\to 1^{-}}{\lim }g\left(x\right)=\underset{x\to 1^{-}}{\lim }{x}^2=1$

and

The left-handed and right-handed limits are equal, and the function value at $x=1$ is also 1. Therefore, $g\left(x\right)$ is actually continuous there, and does not have a jump discontinuity.

Can the product of a function that has a jump discontinuity at $x=3$ and a continuous function be continuous at $x=3$?

Answer:

Sure! Take your continuous function to be the one that is identically zero, or in other words $g\left(x\right)=0$ for all values of $x$. From the previous example, you know that

$f\left(x\right)=\left\{\begin{array}{ll}x+1,& x\le 3\\ x-1,& x>3\end{array}\right.$

has a jump discontinuity at $x=3$, but

The product of the two functions is always equal to 0, so the product is in fact continuous at$x=3$.

Can the product of two functions, both with jump discontinuities at $x=3$, be continuous at $x=3$?

Answer:

You bet! Take your functions to be defined by

$f\left(x\right)=\left\{\begin{array}{ll}1,& x\le 3\\ -1,& x>\hspace{0.17em}3\end{array}\right.$,

and

$g\left(x\right)=\left\{\begin{array}{ll}-1,& x\le 3\\ 1,& x>3\end{array}\right.$.

Both of these functions have a jump discontinuity at $x=3$, but their product is

$f\left(x\right)·g\left(x\right)=\left\{\begin{array}{ll}1·(-1),& x\le 3\\ (-1)·1,& x>3\end{array}\right.=\left\{\begin{array}{ll}-1,& x\le 3\\ -1,& x>\hspace{0.17em}3\end{array}\right.=-1$,

which is continuous at $x=3$.