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When is the following function continuous? How would i go about listing the removable discontinuities and then redefine the function so that it is now continuous in those places?

$$f(x)= \begin {cases} \sqrt{2x}+7&\text{if }x <2\\ 0\ &\text{if }x = 2\\ 3^x&\text{if }x > 2 \end {cases}$$

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    $\begingroup$ That's five questions in an hour. Slow down, have a look at the answers, digest them; then, if you still have questions, come back with one or two more. $\endgroup$ Jun 25, 2012 at 1:52
  • $\begingroup$ Sorry im just trying to learn as much as i can and figure this all out my notes are allover but ill try to limit it thanks :) $\endgroup$
    – soniccool
    Jun 25, 2012 at 2:03

2 Answers 2

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It is discontinuous at $x = 2$. Approaching from the left, the limit is $\sqrt{(2)(2)} + 7 = 9$. Approaching from the right, you have $3^2 = 9$. However at $x = 2$, the function is defined to be $0 \neq 9$. Hence it is discontinuous at $x = 2$, but can be made continuous by defining $f(2) = 9$.

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  • $\begingroup$ So we need f(2)= 9 in order for it to be continuous? $\endgroup$
    – soniccool
    Jun 25, 2012 at 2:38
  • $\begingroup$ @soniccool Yes. $\endgroup$
    – William
    Jun 25, 2012 at 3:46
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Hint: Try drawing a picture of $f(x)$, even if it's just from, say, $x=1$ to $x=3.$ Try to figure out the limit of $f(x)$ to the left and right of $x=2$. Is it the same? What happens when $x=2$?

Remark: If the limit on the left and right hand side is the same (i.e. the function approaches the same $y$-value from both sides of the neighboring $x$ value) , there is a good chance that the function is continuous. Often one defines continuity as "not having to lift your pencil up when you graph the function." However, note that there is indeed a point discontinuity at $x = 2$ due to the limit from the LHS being $9$ and the limit on the RHS being $9$, but the function is defined to be $0$ at $x=2$. This implies a point discontinuity which can be removed if you re-define $f(2)$ to equal $9$ for your original function. This would make $f(x)$ continuous everywhere along the real numbers.

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