Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Working on my AP Calc summer assignment and I am having a hard time understanding how to solve this; I could really use some very dumbed-down help if possible because I don't even know where to start. Here it is...

Determine where the function is continuous and where the function is differentiable.

$$f(x)=\begin{cases}(x+1)^2,& x \leq 0\\ 2x+1,& 0< x < 3\\ (4-x)^2,& x \geq 3\end{cases}$$

Thank you in advance for your help!

share|cite|improve this question
Hint: Is it already clear to you that the function is continuous and differentiable everywhere except possibly 0 and 3, and you simply need to find which properties it fulfills at 0 and 3? – cobaltduck Aug 23 '11 at 17:29
That makes sense but how would I go about doing that? I've never worked with piecewise functions before. – Kaleidoscopic Aug 23 '11 at 17:39

As is said in the comments, everything is clear except at $0$ and $3$.

To see if it is continuous at $0$, for example, you need to check that the definition of continuity at a point is satisfied at $0$. That is, is it true that $\displaystyle \lim_{x\to 0}f(x) = f(0)$?

For differentiability, you again need to check the definition: Does the limit $\displaystyle \lim_{h \to 0} \frac{f(0+h)-f(0)}{h}$ exist?

Similar checks need to be performed for behavior at $3$.

share|cite|improve this answer
I think I understand this better now, thank you! – Kaleidoscopic Aug 23 '11 at 17:51
You're welcome, @Kaleidoscopic ! – wckronholm Aug 23 '11 at 22:18
@Kaleidoscopic if you find an answer to your question useful, then it is good practice to accept an answer. This will serve to let the community know that your question has been resolved. – wckronholm Aug 24 '11 at 15:53

Your Answer


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.