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Study23

On what interval is the function f(x)=x^3-4x^2+5x concave upward? I found the 1st and 2nd derivative, but now what?

  • one year ago
  • one year ago

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  1. Study23
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    \(\ \Huge So far, I have: \) \(\ \Huge f'(x)=3x^2-8x+5, \) \(\ \Huge f"(x)=6x-8. \)

    • one year ago
  2. tcarroll010
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    Where the second derivative is positive, then you have concave upward, but I just saw what you determined for the derivatives, and they're a bit off. Maybe we should work on that part first.

    • one year ago
  3. Study23
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    Okay, so where did I go wrong?

    • one year ago
  4. tcarroll010
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    f'(x) = 3x^2 - 2x - 1. Not sure where where you got the -8 and the 5. Typos? You did the x^3 term correctly, so I have to believe you are familiar with the procedure. derivative of -x^2 is -2x.

    • one year ago
  5. Study23
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    Oh! I'm so sorry but I just realized that the question has two typos, let me fix that... Sorry about that!

    • one year ago
  6. tcarroll010
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    Maybe the equation in the problem is mistyped? If not, then you should go with my first derivative. We can talk about how we got this, if need be.

    • one year ago
  7. Study23
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    Okay, that's the correct f(x). I was looking at a different equation when typing the function...

    • one year ago
  8. tcarroll010
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    You're doing well! Now, just look for those values of x for which the second derivative is positive for concave upward.

    • one year ago
  9. Study23
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    How do I do that?

    • one year ago
  10. Study23
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    So the first derivative doesn't really have much importance in this question? It's the second derivative that's more important?

    • one year ago
  11. tcarroll010
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    A really good way to remember if the second derivative is supposed to be positive or negative for concave up (it's positive you want) is to remember y=x^2 as an example. Point 2: to answer your question, yes, it's the second derivative that determines concavity.

    • one year ago
  12. Study23
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    What does the first derivative determine then?

    • one year ago
  13. Study23
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    And, how do I determine the values for which this function is positive?

    • one year ago
  14. tcarroll010
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    For y=x^2, the second derivative is 2, which is positive, so always concave up. The first derivative does 2 things. One, you have to get that first to get the second derivative, and 2), the first derivative will tell you where the function is increasing and decreasing.

    • one year ago
  15. Study23
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    Oh okay, I get the first derivative know. I'm still stuck on finding values for which the second derivative is positive..

    • one year ago
  16. Study23
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    now*

    • one year ago
  17. tcarroll010
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    For determining where the function is positive, even though this is a cubic, you caught a break, because you can factor out "x" and then you are left with a quadratic for the other factor. If you need more help, I can assist. As for finding values for where the second derivative is positive, set 6x - 8 > 0. 6x > 8. x > 4/3.

    • one year ago
  18. Study23
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    Okay, but for this problem I don't need to know where the values of the original function is positive, correct? Also, would that mean that the interval for which this function is concave up are \(\ \Huge (\frac{4}{3}, +\infty) ?\)

    • one year ago
  19. tcarroll010
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    yes, concave up in that interval. The left "(" should be a "["

    • one year ago
  20. tcarroll010
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    As for whether or not you need to know where the function values are positive, that is needed only if you are told to get them. You don't need to know that to determine concavity.

    • one year ago
  21. Study23
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    Okay! I thought it was a ( because the inequality was greater than not greater than or equal to

    • one year ago
  22. tcarroll010
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    You are understanding this problem well now it seems.

    • one year ago
  23. Study23
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    :)

    • one year ago
  24. tcarroll010
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    As for +/- values, if you are graphing, the equation is + for positive x and - for negative x. Because the quadratic factor is always positive.

    • one year ago
  25. Study23
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    \(\ \Huge Okay, \mathsf T\mathsf h\mathsf a\mathsf n\mathsf k\mathsf s \mathsf f\mathsf o\mathsf r \mathsf a\mathsf l\mathsf l \mathsf t\mathsf h\mathsf e\mathsf h\mathsf e\mathsf l\mathsf p\mathsf ! \)

    • one year ago
  26. tcarroll010
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    You are quite welcome! Nice working with you!

    • one year ago
  27. Study23
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    It was nice working with you, as well :)

    • one year ago
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