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darkigloo

  • one year ago

If f(x)=(x^2)(e^x) then f is decreasing for all x such that A. x<-2 B. -2<x<0 C. x>-2 D. x<0 E. x>0

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  1. myininaya
    • one year ago
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    Calculus or algebra?

  2. darkigloo
    • one year ago
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    calculus

  3. myininaya
    • one year ago
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    Ok cool stuff. So do you know how to find the derivative. The derivative is the function that tells if the original function is decreasing or increasing.

  4. darkigloo
    • one year ago
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    i think the derivative is e^x(x^2+2x)

  5. myininaya
    • one year ago
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    that is right So when is the derivative 0 (this is to find when the function is doing neither increasing or increasing) I'm asking this: solve e^x(x^2+2x)=0

  6. myininaya
    • one year ago
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    e^x is never 0

  7. myininaya
    • one year ago
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    So you are basically looking at solving x^2+2x=0 right now.

  8. darkigloo
    • one year ago
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    x=0 x=-2

  9. myininaya
    • one year ago
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    ok great. Now lets check the intervals around those numbers to see if we have f'>0 (which means f is increasing) or if we have f'<0 (which means f is decreasing) So what I'm asking you to do is look at the interval (-inf,-2) Take a number from this set and see if f'>0 or <0. Then we will also check the set of numbers (-2,0) and also check (0,inf)

  10. myininaya
    • one year ago
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    we have three intervals to check.

  11. myininaya
    • one year ago
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    when ever you plug in a number into e^x all you need to know is that the output will always be positive

  12. myininaya
    • one year ago
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    So again looking at just x^2+2x

  13. myininaya
    • one year ago
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    A number from the group (-inf,-2), we could choose -4. (-4)^2+2(-4)=a positive number>0 so on this interval it is increasing. Now you check (-2,0).

  14. darkigloo
    • one year ago
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    for (-2,0) its negative. for (0,-inf) its positive. So the answer is B. -2<x<0 right!? :D

  15. myininaya
    • one year ago
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    that is right

  16. darkigloo
    • one year ago
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    thanks. and for the e^x...that is always positive?

  17. myininaya
    • one year ago
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    any positive number raised to any power will always be positve

  18. darkigloo
    • one year ago
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    ok. thank you.

  19. myininaya
    • one year ago
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    |dw:1393014251807:dw|

  20. myininaya
    • one year ago
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    Like you can also look at the graph if you want it gets closer to 0 as we approach negative infinity but never touches or goes below

  21. darkigloo
    • one year ago
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    ohh right...i'll have to remember that graph. thanks :)

  22. myininaya
    • one year ago
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    good luck with your calculus

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