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Math Help - Continuity Problems

  1. #1
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    Continuity Problems

    I'm having a heck of a time figuring out how they got an answer to a problem. Maybe you guys can help me out?

    Find values of X, if any at which f is not continuous.

    X/2x^2+x books answer is 0 and -1/2

    Wouldn't I factor this denominator? I understand why its 0 and -1/2 has it makes the bottom denominator equal to 0 but how did they get to that answer? I know just plugging in 0 will prove to make the denominator 0 but how did they do the -1/2? Wouldn't they factor? Thanks

    Another one i'm having problems with is

    radical x-8 with index 3. Don't even know how to start this one.

    THanks
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  2. #2
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    Quote Originally Posted by dan87951 View Post
    I'm having a heck of a time figuring out how they got an answer to a problem. Maybe you guys can help me out?

    Find values of X, if any at which f is not continuous.

    X/2x^2+x books answer is 0 and -1/2

    Wouldn't I factor this denominator? I understand why its 0 and -1/2 has it makes the bottom denominator equal to 0 but how did they get to that answer? I know just plugging in 0 will prove to make the denominator 0 but how did they do the -1/2? Wouldn't they factor? Thanks

    Another one i'm having problems with is

    radical x-8 with index 3. Don't even know how to start this one.

    THanks
    If you mean \frac{x}{2x^2 + x}

    Then, you have a discontinuity whenever the entire denominator = 0.

    That is, when 2x^2 + x = 0

    \implies x(2x + 1) = 0

    \underbrace{x}\underbrace{(2x + 1)} = 0

    Which =0 when either under-bracketed part =0.

    i.e., x= 0 or, when 2x+1 = 0 \implies x=-\frac{1}{2}
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  3. #3
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    Quote Originally Posted by dan87951 View Post
    I'm having a heck of a time figuring out how they got an answer to a problem. Maybe you guys can help me out?

    Find values of X, if any at which f is not continuous.

    X/2x^2+x books answer is 0 and -1/2

    Wouldn't I factor this denominator? I understand why its 0 and -1/2 has it makes the bottom denominator equal to 0 but how did they get to that answer? I know just plugging in 0 will prove to make the denominator 0 but how did they do the -1/2? Wouldn't they factor? Thanks

    Another one i'm having problems with is

    radical x-8 with index 3. Don't even know how to start this one.

    THanks
    Dear Anonymous1,

    \frac{x}{2x^2-x}=\frac{x}{x(2x-1)}=\frac{x}{2x\left(x-\frac{1}{2}\right)}

    Hope this will help you.
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  4. #4
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    I think its just easier for me to use the quadratic formula for this this type of problem. I don't understand the algebra you guys are doing. Is there a website that will help me understand this?
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  5. #5
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    Quote Originally Posted by dan87951 View Post
    I think its just easier for me to use the quadratic formula for this this type of problem. I don't understand the algebra you guys are doing. Is there a website that will help me understand this?
    Dear dan87951,

    What do you don't understand? I have factorized the denominator.

    2x^2-x=x(2x-1); taking x outside the bracket

    x(2x-1)=2x\left(x-\frac{1}{2}\right); taking 2 outside the bracket

    Hope this will help you.
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