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haterofmath

  • 2 years ago

decide if the given function is continuous at the specific value of x...

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  1. haterofmath
    • 2 years ago
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    \[f(x)= \frac{ 2x-4 }{ 3x-2 } at x=2\]

  2. helder_edwin
    • 2 years ago
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    remember the definition: f is continous at x=a if \[ \large \lim_{x\to a}f(x)=f(a) \]

  3. cinar
    • 2 years ago
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    yes it is continuous..

  4. cinar
    • 2 years ago
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    it is defined at x=2 right

  5. cinar
    • 2 years ago
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    f(2)=0 \[\large \lim_{x\to 2}f(x)=0=f(2)\]

  6. cinar
    • 2 years ago
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    and also this is its graph http://www.wolframalpha.com/input/?i=plot+%282x-4%29%2F%283x-2%29 as you can see the func. is continuous at x=2, no hole no infinity

  7. haterofmath
    • 2 years ago
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    ok. so if they were ask to list all the value of x for which the given function is not continuous...how would that work? Example: \[f(x)= \frac{ 3x-1 }{ 2x-6 }\]

  8. cinar
    • 2 years ago
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    the func. is discontinuous when denominator is 0

  9. cinar
    • 2 years ago
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    2x-6=0 x=3 so, for x=3 the func. is not continuous

  10. haterofmath
    • 2 years ago
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    and what about the numerator?

  11. cinar
    • 2 years ago
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    it can be any number, no problem with it.. look at its graph http://www.wolframalpha.com/input/?i=plot+ \frac{+3x-1+}{+2x-6+}

  12. cinar
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=plot+%283x-1%29%2F%282x-6+%29

  13. haterofmath
    • 2 years ago
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    ok, so there's only one value of x?

  14. cinar
    • 2 years ago
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    in this case yes..

  15. haterofmath
    • 2 years ago
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    ok. so how about in this case. f(x) = 3x-2 if x<0 x^2+x if \[x \ge0\]

  16. cinar
    • 2 years ago
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    the func. is not continuous at x=0 why because limit does not exists when we approach to 0, from left limit is -2, from right limit is 0 http://www.wolframalpha.com/input/?i=piecewise+ [{{3x-2%2Cx%3C0}%2C{x^2%2Bx%2Cx%3E%3D0}}]

  17. cinar
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=piecewise [{{3x-2%2Cx%3C0}%2C{x^2%2Bx%2Cx%3E%3D0}}]

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