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anonymous

  • one year ago

find the intervals of increase and decrease for e^2x-e^x, please help!

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  1. Michele_Laino
    • one year ago
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    here we have to compute the first derivative of your function

  2. Michele_Laino
    • one year ago
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    I got this: \[\Large f'\left( x \right) = {e^x}\left( {2{e^x} - 1} \right)\]

  3. Michele_Laino
    • one year ago
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    now the intervals of the real line in which f(x) is increasing, are given by the subsequent condition: \[\Large f'\left( x \right) > 0 \Rightarrow {e^x}\left( {2{e^x} - 1} \right) > 0\]

  4. Michele_Laino
    • one year ago
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    which, since e^x is always positive, is equivalent to this one: \[\Large 2{e^x} - 1 > 0\] please solve that inequality, what do you get?

  5. anonymous
    • one year ago
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    e^x>1/2

  6. Michele_Laino
    • one year ago
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    please now solve it for x

  7. anonymous
    • one year ago
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    would it be ln(e)>1/2

  8. Michele_Laino
    • one year ago
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    no, since if we take the logarithm of both sides, we get: \[\Large \begin{gathered} x\left( {\ln e} \right) > \ln \left( {1/2} \right) \hfill \\ x > - \ln 2 \hfill \\ \end{gathered} \]

  9. Michele_Laino
    • one year ago
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    am I right?

  10. Michele_Laino
    • one year ago
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    please remember that: \[\Large \begin{gathered} \ln e = 1 \hfill \\ \ln \left( {1/2} \right) = \ln 1 - \ln 2 = 0 - \ln 2 = - \ln 2 \hfill \\ \end{gathered} \]

  11. anonymous
    • one year ago
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    oh okay i see it now

  12. Michele_Laino
    • one year ago
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    so your function is increasing in: [-ln 2, +infinity) and decreasing, in: (-infinity, -ln2)

  13. Michele_Laino
    • one year ago
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    the domain of your function is all the real line

  14. anonymous
    • one year ago
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    okay! and to find the local max and min i would have to find the second derivative and plug in -ln2 for x right?

  15. Michele_Laino
    • one year ago
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    no, the local max and min comes from the first derivative

  16. Michele_Laino
    • one year ago
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    now we have this: |dw:1435213594909:dw|

  17. Michele_Laino
    • one year ago
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    as we can see, point x=-ln(2) is a local minimum for f(x)

  18. anonymous
    • one year ago
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    and I plug that into the first derivative so i can get the interval?

  19. Michele_Laino
    • one year ago
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    you can get the intervals in which your function is an increasing or a decreasing function, only solving the inequality: f '(x) >0 for example Now we have found x=-ln(2) or, equivalently e^x= 1/2. At that point the value of your function is: \[\Large f\left( { - \ln 2} \right) = \frac{1}{4} - \frac{1}{2} = - \frac{1}{4}\]

  20. anonymous
    • one year ago
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    so my max would be at (-ln(2),-1/4)?

  21. Michele_Laino
    • one year ago
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    it is a local minimum

  22. anonymous
    • one year ago
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    wait thats the min right?

  23. Michele_Laino
    • one year ago
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    right!

  24. anonymous
    • one year ago
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    and theres no max for this function right?

  25. Michele_Laino
    • one year ago
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    that's right! there are no points of local maximum

  26. anonymous
    • one year ago
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    for my concave up and concave down, would it be for concave down (-ln(2),-1/4) and for concave up (-infinity,-ln(2))U(-ln(2),+infinity)?

  27. Michele_Laino
    • one year ago
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    in order to establish the concavity type, we have to study the sign of the second derivative

  28. Michele_Laino
    • one year ago
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    after a simple computation, I got this: \[\large f''\left( x \right) = {e^x}\left( {4{e^x} - 1} \right)\]

  29. Michele_Laino
    • one year ago
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    now, our function is concave up when f ''(x)>0, and it is concave down when f ''(x)<0 |dw:1435214691996:dw|

  30. Michele_Laino
    • one year ago
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    so we have to solve this inequality: \[\Large f''\left( x \right) > 0 \Rightarrow {e^x}\left( {4{e^x} - 1} \right) > 0\]

  31. Michele_Laino
    • one year ago
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    next, since e^x is always positive, then that inequality is equivalent to this one: \[\Large 4{e^x} - 1 > 0\] please solve it for x, what do you get?

  32. anonymous
    • one year ago
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    would it be x>ln1/4

  33. Michele_Laino
    • one year ago
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    that's right!

  34. Michele_Laino
    • one year ago
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    we can write it as below: \[\Large x > - \ln 4\]

  35. Michele_Laino
    • one year ago
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    so your function, is concave down in this interval: (-infinity, -ln4), and concave up in this interval: [-ln4, +infinity)

  36. anonymous
    • one year ago
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    to find the inflection point I would plug in -ln4 into the second derivative and then that result back into the original equation right?

  37. anonymous
    • one year ago
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    oh wait nvm i would solve for x in the second derivative?

  38. Michele_Laino
    • one year ago
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    no, you have to plug in -ln4 into the function, like this: \[\large f\left( { - \ln 4} \right) = {\left( {\frac{1}{4}} \right)^2} - \frac{1}{4} = \frac{1}{{16}} - \frac{1}{4} = - \frac{3}{{16}}\]

  39. Michele_Laino
    • one year ago
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    so your inflection point is: \[\Large \left( { - \ln 4, - \frac{3}{{16}}} \right)\]

  40. anonymous
    • one year ago
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    oh okay i understand now

  41. anonymous
    • one year ago
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    thank you so much!! I really appreciated it!!

  42. Michele_Laino
    • one year ago
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    :)

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