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$ f(x)= \begin{cases} 4^x&\text{if}\, x\leq 1\\ \frac{9-x^2}{3-x}&\text{if}\, 1<x\leq 4\\ \sqrt x&\text{otherwise} \end{cases} $

(b) at which of these numbers is f continuous from the right, the left, or neither?

(c) sketch the graph of $f$

My question is what does the "otherwise" mean? How do I use it to find the limit on the right or left? and when sketching do I include it?

I already got the number in which $f$ is continuous at $4^x$ and $\frac{9-x^2}{3-x}$. I'm just confused on the $\sqrt x$ one.

Any help is much appreciated.

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  • $\begingroup$ otherwise says something about all $x \in \mathbb{R}$ that do not satisfy the first two constraints. $\endgroup$ – user394255 Feb 27 '17 at 19:16
  • $\begingroup$ In your case, otherwise means that it is going to be $\sqrt{x}$ for any number where $x>4$ since all other parameters from $-\infty$ to 4 are covered, so it will be continuous from any number greater than 4 to infinity as well. $\endgroup$ – Heavenly96 Feb 27 '17 at 19:47
  • $\begingroup$ So for this I calculate the limit as x approaches 4 from the right, correct ? $\endgroup$ – Prisilla Feb 27 '17 at 23:03
  • $\begingroup$ @Prisilla, yes that is what you should do for this one as if you approach it from the left it will just be whatever you got for the middle part of the piece-wise function. $\endgroup$ – Heavenly96 Feb 28 '17 at 16:59

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