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ellieb34

Find all values of x for which the function y=(2(8-5x))/(x+1)

  • one year ago
  • one year ago

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  1. mukushla
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    i think question is incomplete !

    • one year ago
  2. ellieb34
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    is differentiable.

    • one year ago
  3. ellieb34
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    y=([\sqrt{8-5x}\])/(x+1)...it's a square root also..not a two

    • one year ago
  4. mukushla
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    \[f(x)=\frac{\sqrt{8-5x}}{x+1}\]

    • one year ago
  5. ellieb34
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    yes, that's what it looks like!

    • one year ago
  6. mukushla
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    ok u must check for continuity first

    • one year ago
  7. ellieb34
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    when I graphed it, it looked like 1/x graph

    • one year ago
  8. mukushla
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    yeah its something like that..maybe

    • one year ago
  9. ellieb34
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    so how do I check for continuity?

    • one year ago
  10. mukushla
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    ok u have some restrictions here.. division by zero is not allowed under radical cant be negative

    • one year ago
  11. mukushla
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    actually u must find domain of ur function

    • one year ago
  12. ellieb34
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    domain is x=-1

    • one year ago
  13. mukushla
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    \[x\neq -1\]

    • one year ago
  14. ellieb34
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    ohyeah you're right

    • one year ago
  15. mukushla
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    and what about radical?

    • one year ago
  16. ellieb34
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    what's radical?

    • one year ago
  17. mukushla
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    square root

    • one year ago
  18. ellieb34
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    x\[\neq5/8\]

    • one year ago
  19. ellieb34
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    er 8/5

    • one year ago
  20. mukushla
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    when u have something like \[\sqrt{x}\] u must have \[x\ge0\] right ?

    • one year ago
  21. ellieb34
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    yes

    • one year ago
  22. mukushla
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    so what about the expression under square root sign?

    • one year ago
  23. ellieb34
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    \[x \neq 8/5\]

    • one year ago
  24. mukushla
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    emm i mean u must have \(8-5x\ge0\) and this leads to\[x\le\frac{8}{5}\]make sense?

    • one year ago
  25. ellieb34
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    oh yes sorry i meant to do that sign instead of the nonequal sign

    • one year ago
  26. ellieb34
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    it does make sense

    • one year ago
  27. mukushla
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    good

    • one year ago
  28. mukushla
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    ok so the domain of function becomes\[x\le \frac{8}{5} \ \ \text{and} \ \ x\neq-1\]

    • one year ago
  29. mukushla
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    or u can write it like this\[(-\infty,-1)\cup (-1,\frac{8}{5}]\]

    • one year ago
  30. ellieb34
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    got it. so is that where it's differentiable?

    • one year ago
  31. mukushla
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    yes thats it ... and just one more point function is not continous at end points \(x=-1\) and \(x=\frac{8}{5}\) so it'll not be differentiable at that points...

    • one year ago
  32. ellieb34
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    ok thank you so much! i appreciate it :)

    • one year ago
  33. mukushla
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    very welcome :)

    • one year ago
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