A function that has both removable and jump discontinuity. I want to draw a function that has a removable discontinuity at x=1 and jump discontinuity at x=3. I figured the following function:
x+(x+1)/(x-1)+(x-3)

My rationale is that it gives removable at (x-1), jump at (x+1)/(x-3). However, I am not sure if my process is correct.I have been ignoring calculus for long but now decided to learn down cold, with the help of this site maybe! 
 A: There are many functions which will have a removable discontinuity at $x=1$ and a jump discontinuity at $x=3$. 
Generally for a removable discontinuity at $x=a$ you need a term such as $\dfrac{x-a}{x-a}$ in your function. 
For a jump discontinuity at an integer, the floor function $\lfloor x\rfloor$ does nicely. A term such as $\lfloor x\rfloor(x-n)$ has a jump discontinuity everywhere except at the integer $n$. 
So the following function has a removable discontinuity at $x=1$ and a jump discontinuity at every integer except $1$:
\begin{equation}
f(x)=\dfrac{x-1}{x-1}+\lfloor x\rfloor(x-1)
\end{equation}
It is undefined at $x=1$.
On the interval $[0,1)$, $f(x)=1$.
On the interval $(1,2)$, $f(x)=1+1\cdot(x-1)=x$.
On the interval $[2,3)$, $f(x)=1+2\cdot(x-1)=2x-1$. Since $f(x)$ approaches $2$ as $x$ approaches $2$ from the left but $f(2)=3$ there is a jump discontinuity at $x=2$. Likewise at $x=3$ and every other integer with the exception of $x=1$.
On the interval $[3,4)$, $f(x)=1+3\cdot(x-1)=3x-2$ so $f(3)=7$ but $\lim_{x\to3^-}f(x)=5$ so there is a jump discontinuity at $x=3$
Generally, for a removeable discontinuity at an integer $x=n$ and a jump discontinuity at every other integer the following function will do:
\begin{equation}
f(x)=\dfrac{x-n}{x-n}+\lfloor x\rfloor(x-n)
\end{equation}
A: I accepted the earlier answer but it's hard to understand and it took me time.So I thought although the earlier answer is elegant and full, I will write a simple answer that anybody can understand.
I learned that most jump discontinuous functions are piece wise functions. They will have two different set of y values for two different set of x values and in the graph it will look like the function jumps from its previous region.

In the above graph you can see that we have a quadratic and a linear function.The graph looks like jumps down from up above.One thing that is important to note that it is the same function defined in two different regions and hence should not be mistaken for two different functions, rather two pieces of the same function.
Looking at the graph I can say that the function is undefined at x=2 when it's quadratic when x<2.So we can say one piece would be x² when x <2.
The other piece would be defined at 2 and simply y=1 when x>=2.
