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Example 7 (Cont,Removable DisCont) For a function f (x), Ifind the intervals of continuity (IC); II. determine the removability of each DisCont III: for each remova...

Question

Example 7 (Cont,Removable DisCont) For a function f (x), Ifind the intervals of continuity (IC); II. determine the removability of each DisCont III: for each removable discontinuity at x = &, define a new function g(x) continuous at x=a and equal to f (x) elsewhere. Complete I-III for each function f below. (a) f(t)=-7-614+97 (b) f(t)=e " 672+91 (c) f(t) =el?-5t+6) -$ (d) f(x) = Inkx?) V64-r (e) f(x)=13 In(x+3) (f) f(0) =Osin (8) (g) f(0) =&sin(0)

Example 7 (Cont,Removable DisCont) For a function f (x), Ifind the intervals of continuity (IC); II. determine the removability of each DisCont III: for each removable discontinuity at x = &, define a new function g(x) continuous at x=a and equal to f (x) elsewhere. Complete I-III for each function f below. (a) f(t)=-7-614+97 (b) f(t)=e " 672+91 (c) f(t) =el?-5t+6) -$ (d) f(x) = Inkx?) V64-r (e) f(x)=13 In(x+3) (f) f(0) =Osin (8) (g) f(0) =&sin(0)



Answers

Find all values $x=a$ where the function is discontinuous. For each point of discontinuity, give (a) $f(a)$ if it exists, (b) $\lim _{x \rightarrow a} f(x),(c) \lim _{x \rightarrow a^{+}} f(x),(\text { d }) \lim _{x \rightarrow a} f(x),$ and (e) identify which conditions for continuity are not met. Be sure to note when the limit doesn't exist.

So, based on the graph here, let's observe. So have X is minus one. There seems to be some changes there. So let's observe f of minus one. We have ah solid dot at two so f of minus one is to Now let's observe f of minus one from the left side off of minus one minus. So it looks like it is too. However, if we observe ah, the limit of f of X as it was a limit here as X goes to minus one plus, this one does not exist and your limit as f of X as X goes to minus one also does not exist. So since your limit does not exist there, so he here since limit, um f of X as X goes to minus one, does not exist. Your function is discontinuous

So let's observe here for number two. So part a start with negative one so f of negative one that's equal to to All right, So for part B, now let's find limit of F of X as X goes to minus one minus. Looks like it's to sequel it, too now for limit of F of X as X goes to minus one. Plus looks like that's equal to four. Finally, let's observe the limit f of X as X goes toe, one minus one does not exist. So since that's the case, so e since limit of f of X as X goes toe, one does not exist minus one. Here, your function is discontinuous.

Let's up there of this function here at one. So let's see if f of one exists. So there's f of one exists. Yes, it does. We have a solid dot at two, so f of one is equal to two. Let's see part B. So limit f of X as X goes to one minus. Looks like it's negative, too. Let's see the limit of F of X as X goes to one plus. So the right side of one that's also minus two. So in this case, limit f of X as X goes to one is equal to minus two. Now all this stuff exists. However, those two don't match here. So this doesn't match this. And for that reason, so since F of one is not equal to limit f of X as X approaches one, your function is discontinuous.

So we have to critical values here. So X is three and X is minus two. Let's observe things for X is three so f of three. It looks like it's equal to one No for part B and limit as X goes to one minus, um three minus. That's gonna be minus one that's equal to minus one and the same thing with limit as X goes to three plus. So the right side of three it all. It also goes to minus one, so the limit as X goes to three. Is it called a minus one. So in this case, we have effort. Three is not equal to limit. X goes to three of f of X so discontinuous you can do the same Procedure for X is equal to minus two, so for X is minus two as well, So f of minus two is equal to one limit. F of X is X goes to minus two minus is equal to minus one as well. Same thing here Limit of f of X as X goes to minus to plus his minus one. So limit of f of X as X goes to minus two is equal to minus one. However, F of minus two is not equal to limit of F of X as X goes to minus two. Since that's the case, it's discontinuous.


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