My Math Forum First Post... Continuity of Piecewise func

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February 21st, 2018, 12:23 PM   #1
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First Post... Continuity of Piecewise func

I know this is probably pretty basic, but just started in Calculus. Any help would be appreciated. I know the limit from the left is 3, and the limit from the right is 3, and the f(1)=4 which makes is discontinuous. Not sure how to format the interval to make it continuous.
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 February 21st, 2018, 12:54 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,299 Thanks: 1688 Moved to Pre-Calculus. The function is undefined at x = 0 (or -2 if the denominator was intended to be x+2) and discontinuous at x = 1. What notation for intervals have you been taught? Last edited by skipjack; February 21st, 2018 at 03:53 PM. Reason: to correct, as I'd misread the image
 February 21st, 2018, 03:36 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 The function I see (perhaps it has been changed?) is f(x)= 9/(x+ 2) if x< 1 (you wrote "9/x+ 2" but I think that 9/(x+ 2) was intended) f(x)= 4 if x= 1 and f(x)= 5x- 2 if x> 1 The limit at 1 "from the left" is the limit of 9/(x+ 2). That function is continuous at x= 1 so the limit is 9/(1+ 2)= 9/3= 3. The limit "from the right" is the limit of 5x- 2. That function is continuous at x= 1 so the limit is 5(1)- 2= 3. That is enough to say that "the limit" is 3. But the value of the function is 4 at x= 1 so, even though the limit exists, the function is NOT continuous at x= 1.
 February 21st, 2018, 06:32 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 I suspect that the question is mis-stated. I suspect the real question is how to define the function to make it continuous at x = 1.
February 22nd, 2018, 08:00 AM   #5
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Quote:
 Originally Posted by Country Boy The function I see (perhaps it has been changed?) is f(x)= 9/(x+ 2) if x< 1 (you wrote "9/x+ 2" but I think that 9/(x+ 2) was intended) f(x)= 4 if x= 1 and f(x)= 5x- 2 if x> 1 The limit at 1 "from the left" is the limit of 9/(x+ 2). That function is continuous at x= 1 so the limit is 9/(1+ 2)= 9/3= 3. The limit "from the right" is the limit of 5x- 2. That function is continuous at x= 1 so the limit is 5(1)- 2= 3. That is enough to say that "the limit" is 3. But the value of the function is 4 at x= 1 so, even though the limit exists, the function is NOT continuous at x= 1.
Okay, so redefining f(1)= 3 rather than 4 makes the function continuous.

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