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Math Help - Continuity!

  1. #1
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    Continuity!

    f(x) = { -x+b if x<1 or 1 if x=1 or (-5/(x-b))-4 if x>1 and x doesnot equal b}

    (a) For what value(s) of b is f continuous at 1 ?

    (b) For what value(s) of b does f have a removable discontinuity at 1 ?

    (c) For what value(s) of b does f have an infinite discontinuity at 1 ?

    (d) For what value(s) of b does f have a (finite) jump discontinuity at 1 ? Write your answer in interval notation.

    I literally have no idea where to start. If someone could just atleast get me started it would be a huge help. Thanks.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by demonmath View Post
    f(x) = { -x+b if x<1 or 1 if x=1 or (-5/(x-b))-4 if x>1 and x doesnot equal b}

    (a) For what value(s) of b is f continuous at 1 ?

    (b) For what value(s) of b does f have a removable discontinuity at 1 ?

    (c) For what value(s) of b does f have an infinite discontinuity at 1 ?

    (d) For what value(s) of b does f have a (finite) jump discontinuity at 1 ? Write your answer in interval notation.

    I literally have no idea where to start. If someone could just atleast get me started it would be a huge help. Thanks.
    f(x)=\left\{\begin{array}{lr}x+b&:x<1\\1&:x=1\\ \frac{-5}{x-b}-4&:x>1\end{array}\right\}

    (a) f(x) is continuous at 1 if \lim_{x\to1^-}f(x)=f(1)=\lim_{x\to1^+}f(x)

    So you want both limits to equal f(1), which is 1.

    For the left limit, we use f(x)=-x+b. The limit as x\to1^- is -1+b, so b needs to be 2. Now, since the other equation for x>1 also depends on b, we have to hope that b=2 works for it as well. And it does: \lim_{x\to1^+}f(x)=\frac{-5}{1-2}-4=5-4=1, so b=2 implies f is continuous at x=1.

    (b) Here you want \lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)\neq 1 (and also \neq\pm\infty).

    (c) Here you want the function to blow up near x=1, so what value of b can you choose so that f\to\infty or f\to-\infty as x\to1?

    (d) Here you want \lim_{x\to1^-}f(x)\neq\lim_{x\to1^+}f(x)\neq\pm\infty. This is basically a "what's left over" question. Any value of b that was not an answer to one of the above parts will work here.
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  3. #3
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    Wow I definitely understood that first one when you walked me through it thanks a ton. I don't know how to continue on though, I thought I just needed a starting point but clearly that's not enough because I am still struggling with the basic concept there
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by demonmath View Post
    Wow I definitely understood that first one when you walked me through it thanks a ton. I don't know how to continue on though, I thought I just needed a starting point but clearly that's not enough because I am still struggling with the basic concept there
    For (b), we want to solve

    \lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x) i.e. -1+b=\frac{-5}{1-b}-4

    -1+b=\frac{-5}{1-b}-4\implies b+3=\frac{-5}{1-b}\implies (b+3)(1-b)= -5\implies -b^2-2b+3=-5\implies b^2+2b-8\implies(b+4)(b-2)=0\implies b=\{-4,2\}

    However, we have stated that for this part these limits cannot equal 1, and from part (a) we know that b=2 means that they do, so b\neq2. Thus, only b=-4 will give you a removable discontinuity (a "hole"), because \lim_{x\to1^-}(-x-4)=-5=\lim_{x\to1^+}\frac{-5}{x+4}-4.

    For part (c) we just want the denominator approach 0 as x\to1^+, so b=1 will do the trick by making f(x)\to-\infty as x\to1^+.

    For (d), every other value of b will give a finite discontinuity, so the set B=\{b\in\mathbb{R}:b\neq-4,b\neq1,b\neq2\} contains the values you're looking for.
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  5. #5
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    I really appreciate all the help it was really good for my understanding of the material.
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  6. #6
    Super Member redsoxfan325's Avatar
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    You're welcome.
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