Continuous Function

  • #1
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If you are given:

[tex] f(x) = \left\{\begin{array}{cc} (3x)/ (x+4),&\mbox{ if } x\leq 4 , x != -4 & \sqrt{x-3}, &\mbox{ if } x > 4[/tex] (3x is numerator and x+4 is denominator)

find the points of discontinuity of the function. Identify each type. Since [tex] x != 4 [/tex](not equal to 4, don't know how to put not equal sign) , then this is a point of discontinuity. Hence is this a removavble discontinuity (the only discontinuity?) because you get find an extended function?

Thanks a lot

PS: Is there any good LaTex tutorial. As you can see I need to work on it!
 
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Answers and Replies

  • #2
dextercioby
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I think it has 2 points of discontinuity.First is '-4',obviously.

Find the second.

Daniel.

BTW,i think you meant [tex] x\neq -4 [/tex]
 
  • #3
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[tex] x = 3 [/tex] the other one because of domain restrictions?
 
  • #4
dextercioby
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Okay,are u guessing???U have to be sure of your answer...Why is the function not continuous in "3"??

Daniel.
 
  • #5
HallsofIvy
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You say "Since x!= 4 (not equal to 4, don't know how to put not equal sign) , then this is a point of discontinuity." but the formula does NOT say x can't be 4, it says x can't be -4. You may be confusing 4 and -4.
 
  • #6
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3 things must exist for a function to be continuous at a point

(a) lim f(x) exists
x--> a

(b) lim f(x) = f(a)
x-->a

(c) f(a) is defined

So

lim f(x) does not exist. Hence 4 is a point discontinuity
x--> 4

lim f(x) does not exist. Hence -3 is a point discontinuity.
x--> -4

Is this right?

Thanks
 
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  • #7
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any help is appreciated

thanks
 
  • #8
learningphysics
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courtrigrad said:
3 things must exist for a function to be continuous at a point

(a) lim f(x) exists
x--> a

(b) lim f(x) = f(a)
x-->a

(c) f(a) is defined

So

lim f(x) does not exist. Hence 4 is a point discontinuity
x--> 4

lim f(x) does not exist. Hence -3 is a point discontinuity.
x--> -4

Is this right?

Thanks

You're correct that 4 is a point of discontinuity since limx->4 does not exist.

For the second part, I think you meant, hence -4 is a point of discontinuity? You're right that the limx->-4 does not exist, but also f(-4) is undefined. So either reason gives you that there's a discontinuity at x=-4.

Yes, I find according to the question that there are 2 points of discontinuity, 4 and -4.
 
  • #9
dextercioby
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Hey,points o discontinuity are in general "x" values.Tell me why the "x=-3" is a point of discontinuity...

You've already been answered to the problem...

Daniel.
 

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