- #1
courtrigrad
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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!
[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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