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anonymous

  • one year ago

solve the inequality? x^2e^xlnx>0

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  1. anonymous
    • one year ago
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    \[x^2 e^x lnx >0\]

  2. freckles
    • one year ago
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    you only need to solve ln(x)>0

  3. anonymous
    • one year ago
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    isn't x undefined then?

  4. freckles
    • one year ago
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    no... \[\ln(x)=y \implies e^{y}=x \\ \text{ here we have } y=0 \\ \text{ so we have } e^{0}=x\] so how do you solve the equation ln(x)=0

  5. freckles
    • one year ago
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    like what is the solution

  6. freckles
    • one year ago
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    and remember y=ln(x) looks like this: |dw:1440878660198:dw|

  7. anonymous
    • one year ago
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    well e^0 is 1

  8. freckles
    • one year ago
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    drawing got cut off but that line is suppose to be y=0

  9. freckles
    • one year ago
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    so x>1

  10. freckles
    • one year ago
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    x^2e^xln(x)>0 when x>1

  11. anonymous
    • one year ago
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    so in interval notation, is it \[(1,\infty)\]

  12. freckles
    • one year ago
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    yep

  13. anonymous
    • one year ago
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    thank you so much

  14. freckles
    • one year ago
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    do you understand how we got that though ?

  15. freckles
    • one year ago
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    |dw:1440878804304:dw|

  16. freckles
    • one year ago
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    e^x was always positive x^2 was always positive ln(x) was going to determine whether or not e^x*x^2*ln(x) was positive or not

  17. anonymous
    • one year ago
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    yeah, I'm actually kind of mad at myself that I didnt change it to exponential notation.

  18. freckles
    • one year ago
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    just for fun x^2e^xln(x)<0 has solution (0,1)

  19. freckles
    • one year ago
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    I didn't include anything to the left of 0 because ln(x) is undefined there

  20. freckles
    • one year ago
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    left of 0 or 0*

  21. dinamix
    • one year ago
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    |dw:1440879090764:dw| your graph for function is like that so f(x)>0 in ] 1; +infinity [ @hailbug

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