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!
Best Answer
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}