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ellieb34

  • 3 years ago

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

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  1. mukushla
    • 3 years ago
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    i think question is incomplete !

  2. ellieb34
    • 3 years ago
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    is differentiable.

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

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

  5. ellieb34
    • 3 years ago
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    yes, that's what it looks like!

  6. mukushla
    • 3 years ago
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    ok u must check for continuity first

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

  8. mukushla
    • 3 years ago
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    yeah its something like that..maybe

  9. ellieb34
    • 3 years ago
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    so how do I check for continuity?

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

  11. mukushla
    • 3 years ago
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    actually u must find domain of ur function

  12. ellieb34
    • 3 years ago
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    domain is x=-1

  13. mukushla
    • 3 years ago
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    \[x\neq -1\]

  14. ellieb34
    • 3 years ago
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    ohyeah you're right

  15. mukushla
    • 3 years ago
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    and what about radical?

  16. ellieb34
    • 3 years ago
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    what's radical?

  17. mukushla
    • 3 years ago
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    square root

  18. ellieb34
    • 3 years ago
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    x\[\neq5/8\]

  19. ellieb34
    • 3 years ago
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    er 8/5

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

  21. ellieb34
    • 3 years ago
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    yes

  22. mukushla
    • 3 years ago
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    so what about the expression under square root sign?

  23. ellieb34
    • 3 years ago
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    \[x \neq 8/5\]

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

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

  26. ellieb34
    • 3 years ago
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    it does make sense

  27. mukushla
    • 3 years ago
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    good

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

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

  30. ellieb34
    • 3 years ago
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    got it. so is that where it's differentiable?

  31. mukushla
    • 3 years ago
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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...

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

  33. mukushla
    • 3 years ago
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    very welcome :)

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