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mathmath333

  • one year ago

Find the range of \(x\).

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \dfrac{(x+3)(x^2-4)}{(x-1)^2}\leq 0\hspace{.33em}\\~\\ \end{align}}\)

  2. geerky42
    • one year ago
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    Since \((x-1)^2\) is always positive, Multiply both sides by that, then we are down to \((x+3)(x^2-4)\le0\)

  3. geerky42
    • one year ago
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    Note that \(x\neq1\)

  4. imqwerty
    • one year ago
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    1<x<=2

  5. geerky42
    • one year ago
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    Then try find root of \((x+3)(x^2-4)=0\), then determine which intervals between roots exclusive would have \((x+3)(x^2-4)\) be smaller than 0

  6. geerky42
    • one year ago
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    Yeah but there are more @imqwerty

  7. idku
    • one year ago
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    at first \(x\ne1\) now, for this right side to be =0, x=2, -2, -3 for the right side to be greater than 0, (x+3) and (x^2-4) , both have to be negative, or both positive NOTE: (x-1)^2 is always positive x+3 and (x-2)(x+2) are both pos. when x>-2 x+3 and (x-2)(x+2) are both neg. when x<-3 x>-2 or x<-3 and x = 2, -2, -3 CONCLUSION x\(\le\)-3, or x \(\ge\) -2 and x=2

  8. campbell_st
    • one year ago
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    well if you multiply both sides of the equation by (x -1)^2 then you get \[(x + 3)(x^2-4) \le 0\] so this becomes \[(x + 3)(x-2)(x+2) \le 0\] so the zeros are x= -3, -2 and 2 so pick some test values that are between the zeros and check |dw:1433794059127:dw| x = 0 you get -12/1 pick x = -2.5 you get a positive

  9. campbell_st
    • one year ago
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    so the graph would look like |dw:1433794220437:dw| in some way... hope it makes sense

  10. campbell_st
    • one year ago
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    now you can define the intevals where the inequality holds true.

  11. campbell_st
    • one year ago
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    there is an oblique asymptote which won't affect the solution

  12. imqwerty
    • one year ago
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    x<=3 or -2<=x<1 or 1<x<=2

  13. geerky42
    • one year ago
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    Direct answer isn't allowed... @imqwerty

  14. campbell_st
    • one year ago
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    but it is good that you adjusted your answer @imqwerty after seeing some solutions

  15. imqwerty
    • one year ago
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    :)

  16. campbell_st
    • one year ago
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    but then again @imqwerty I think you solution is still incorrect... Can't see how 1 plays a roll in the solution

  17. imqwerty
    • one year ago
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    1 can't be the solution because then the denominator will become 0 @campbell_st

  18. campbell_st
    • one year ago
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    good point

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