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anonymous

  • one year ago

please help! Domain and range of: a. f(x)= 1/x+3 b. g(x)= sqrt(x+6)

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  1. anonymous
    • one year ago
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    also h(x)= x^3+2x+5

  2. rishavraj
    • one year ago
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    is it \[\frac{ 1 }{ x + 3 }~~~~~or ~~~~~~\frac{ 1 }{ x } + 3\]

  3. rishavraj
    • one year ago
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    if its \[\frac{ 1 }{ x + 3 }\] then the denominator can't be zero. i.e.\[x + 3 \neq 0\] so \[x \neq -3\]

  4. anonymous
    • one year ago
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    ok thanks! what about range? and the other ones

  5. anonymous
    • one year ago
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    @rishavraj

  6. rishavraj
    • one year ago
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    did u get the domain ????

  7. anonymous
    • one year ago
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    for which one?

  8. anonymous
    • one year ago
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    i got the first one! @rishavraj

  9. anonymous
    • one year ago
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    and i think i got sqrt(x+3) but i'm not sure

  10. rishavraj
    • one year ago
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    the functn in sqrt can't be negative.... so \[x + 3 \ge 0\]

  11. anonymous
    • one year ago
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    i got x≥-3

  12. anonymous
    • one year ago
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    oops the question was actually sqrt (x+6) so x≥-6 ?

  13. anonymous
    • one year ago
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    or is it x>6? because couldn't it be sqrt(6-6)

  14. rishavraj
    • one year ago
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    its \[x \ge -6\]

  15. anonymous
    • one year ago
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    oh oops i meant -6, but still why isn't it x>-6

  16. anonymous
    • one year ago
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    you looking for the range of \[f(x)=\frac{1}{x+3}\]?

  17. anonymous
    • one year ago
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    yes please

  18. anonymous
    • one year ago
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    i got all real numbers but not sure

  19. anonymous
    • one year ago
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    a fraction is only zero if the numerator is zero

  20. anonymous
    • one year ago
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    therefore \[\frac{1}{x+3}\] cannot be zero because the numerator is 1

  21. anonymous
    • one year ago
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    range is all real numbers except zero

  22. anonymous
    • one year ago
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    oh! oops. thanks! what about sqrt(x+6) and h(x)=x^3+2x+5

  23. anonymous
    • one year ago
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    anyone?

  24. rishavraj
    • one year ago
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    its clear the its domain and range is ALL REAL Numbers... i.e R

  25. anonymous
    • one year ago
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    which one are we talking about :/

  26. anonymous
    • one year ago
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    \[\sqrt{x+6}\] the square root of anything is never less that zero range \[y\geq 0\]

  27. rishavraj
    • one year ago
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    about h(x) = x^3 + 2x + 5

  28. anonymous
    • one year ago
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    cube goes from \(-\infty\) to \(+\infty\)

  29. anonymous
    • one year ago
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    oh, ok thanks! and im still confused about the sqrt one?

  30. anonymous
    • one year ago
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    \[y\geq 0\]

  31. anonymous
    • one year ago
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    why isn't it y>0?

  32. rishavraj
    • one year ago
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    bcoz it can be even zero thts why.....

  33. anonymous
    • one year ago
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    oh. why is the domain x≥-6

  34. anonymous
    • one year ago
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    i thought it could be sqrt0 or no

  35. rishavraj
    • one year ago
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    see the function in sqrt can never be negative...... so \[x + 6 \geq 0\] \[x \geq -6\]

  36. anonymous
    • one year ago
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    oh okay thank you so much!

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