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ksaimouli

  • 5 years ago

find the domain of the function f.use limits to describe the behaviour of f at valus of x not in its domain 1/x+3

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  1. ksaimouli
    • 5 years ago
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    plz explain me

  2. slaaibak
    • 5 years ago
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    For what value of x will you divide by 0?

  3. ksaimouli
    • 5 years ago
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    ?

  4. ksaimouli
    • 5 years ago
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    \[1\div x+1\]

  5. ksaimouli
    • 5 years ago
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    |dw:1327444420449:dw|

  6. slaaibak
    • 5 years ago
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    Cool. \[f(x) = {1 \over x+3}\] Where do you think will this function have a discontinuity? Aka, for what values of x is this function undefined?

  7. ksaimouli
    • 5 years ago
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    -3

  8. slaaibak
    • 5 years ago
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    Yes. So, do you agree this function is defined for every value of x, except for -3?

  9. ksaimouli
    • 5 years ago
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    yup

  10. ksaimouli
    • 5 years ago
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    so ow to write end bhaviour

  11. ksaimouli
    • 5 years ago
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    i am stuck with that

  12. slaaibak
    • 5 years ago
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    Then we can agree that the domain of the function is: \[(-\infty,-3) \cup (-3, \infty)\]

  13. slaaibak
    • 5 years ago
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    That means x goes from -infinity to JUST before three, meaning every value smaller than -3. and then it unites with every value larger than -3 to infinity

  14. ksaimouli
    • 5 years ago
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    thx slaaibak

  15. ksaimouli
    • 5 years ago
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    thank u ver much

  16. slaaibak
    • 5 years ago
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    For the limits: \[\lim_{x \rightarrow -3^-} { 1 \over x + 3} = -\infty\] similarly \[\lim_{x \rightarrow -3^+} { 1 \over x + 3} = \infty\] and lastly \[\lim_{x \rightarrow -3} { 1 \over x + 3} = Does NotExist\]

  17. ksaimouli
    • 5 years ago
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    ok got u but what happens in this case |dw:1327445191194:dw|

  18. ksaimouli
    • 5 years ago
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    i got it -2 right

  19. slaaibak
    • 5 years ago
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    Whichever x makes the denominator zero would make the undefined for that x. \[(-\infty, -2) \cup (-2,2) \cup (2, \infty)\] That's the domain

  20. slaaibak
    • 5 years ago
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    if x^2 - 4 = 0 x = -2 or x = 2

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