Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is How do I graph a continuous function when I am only given restrictions? For example, Sketch the graph of a function that is continuous everywhere except at x=3, and is continuous from the left at x=3. I have another problem to solve, but if I can get help with this one, I am sure I will get the other one as well.

share|improve this question
add comment

2 Answers

The first thing to realize is that there are infinitely many different correct answers. Instead of drawing a sketch, I’ll give a couple explicitly. Let’s start with a function $f(x)$ that’s continuous at every real number. It doesn’t matter what $f$ is, so let’s make it very simple: $f(x)=0$ for all $x\in\Bbb R$. Now we want to make it discontinuous at $x=3$; one simple way to do that is to make it jump at $x=3$. For instance, we could set

$$g(x)=\begin{cases} 0,&\text{if }x<3\\ 1,&\text{if }x>3\\ ?,&\text{if }x=3\;. \end{cases}$$

This function $g$ will certainly be discontinuous at $x=3$ no matter how we fill in the question mark, because

$$\lim_{x\to 3^-}g(x)=0\quad\text{ and }\quad\lim_{x\to 3^+}g(x)=1\;;$$

if $g$ were continuous at $x=3$, the two one-sided limits would have to be equal.

Now what about $g(3)$? We want $g$ to be continuous from the left at $x=3$, which means that we want

$$g(3)=\lim_{x\to 3^-}g(x)=0\;.$$

Thus, the desired function is

$$g(x)=\begin{cases} 0,&\text{if }x\le 3\\ 1,&\text{if }x>3\;. \end{cases}$$

Note that I could have pulled the same trick with any continuous function $f$ by defining

$$g(x)=\begin{cases} f(x),&\text{if }x\le 3\\ f(x)+1,&\text{if }x>3\;. \end{cases}$$

share|improve this answer
add comment

In order to sketch such a function, you need to come up with a function with this property.

I suggest the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by

$f(x) = \begin{cases} 1 & \quad x \leq 3 \\ -1 & \quad x > 3 \end{cases}$

It is clear that the function is continuous everywhere except $3$. It is continuous from the left since $\lim_{x \rightarrow 3; x\leq 3} f(x) = 1 = f(3)$. It is not continuous at the $3$ since the right limit $\lim_{x \rightarrow x; x > 3} f(x) = -1$.

share|improve this answer
    
Actually, one can produce a sketch without having an explicit function. Just draw something that’s continuous on $(\leftarrow,3]$ and $(3,\to)$ and has a jump at $x=3$; this is typically indicated by making a heavy dot at the righthand end of the left half of the graph and an open circle at the lefthand end of the right half. –  Brian M. Scott Jul 14 '12 at 19:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.