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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.
Attached Images piecewise.jpg (17.9 KB, 21 views) February 21st, 2018, 12:54 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,654 Thanks: 2087 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: 3,264 Thanks: 902 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: 1,310 Thanks: 551 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. Tags continuity, func, piecewise, post Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Leesasomething Calculus 2 April 20th, 2011 11:30 PM peaceofmind Calculus 12 April 19th, 2011 12:09 AM Ibanez Calculus 2 April 13th, 2009 03:45 AM Jazzmazz Algebra 2 December 16th, 2007 12:17 PM peaceofmind Algebra 12 December 31st, 1969 04:00 PM

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