# Piecewise Function Calc Confusion

Help me understand this question.

Consider The Function

$$f(x)= \begin {cases} \dfrac{2x-2}{x-1}&\text{if }x \leq2\\ \dfrac{8}{x}\ &\text{if }x \in_\ (2,4)\\ \sqrt x&\text{if }x \geq 4 \end {cases}$$

Where is the function continuous? If there are any removeable discontinuities, list them and then redeﬁne the function so that is now continuous in those places.

What exactly is the question asking for and how do i solve this? Or explain it on a test?

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What are your thoughts about this? Did you check already what happens at the leap points From the right, from the left...limits? –  DonAntonio Jun 18 '12 at 3:59

You are given a function, which is defined in different ways depending on where the input is.

You will want to first consider each part of the function separately; on $(-\infty,2]$, is it always defined? Is it continuous there? When you are looking at the part of the function, all you need to worry about is the formula $$f(x) = \frac{2x-2}{x-1}.$$

For instance, you'll notice that the function is not defined at $x=1$. So it cannot be continuous there; everywhere else it is continuous (quotient of two continuous functions, denominator not equal to $0$). What is happening at $x=1$? Is it a removable discontinuity, or not?

Then consider what happens on $(2,4)$. Then what happens on $[4,\infty)$.

Once you are done considering each piece separately, you need to think about how the pieces fit together. What happens at $x=2$? The function is defined one way on the left, and a different way on the right. Do the two piece fit together properly (that is, are the limits from the right and from the left the same?) If so, what does that tell you about continuity? If not, what does that tell you?

You'll want to then do the same thing at $x=4$.

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