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Study23 Group Title

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?

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. tcarroll010 Group Title
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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.

    • 2 years ago
  3. Study23 Group Title
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    Okay, so where did I go wrong?

    • 2 years ago
  4. tcarroll010 Group Title
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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.

    • 2 years ago
  5. Study23 Group Title
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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!

    • 2 years ago
  6. tcarroll010 Group Title
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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.

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

    • 2 years ago
  8. tcarroll010 Group Title
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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.

    • 2 years ago
  9. Study23 Group Title
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    How do I do that?

    • 2 years ago
  10. Study23 Group Title
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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?

    • 2 years ago
  11. tcarroll010 Group Title
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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.

    • 2 years ago
  12. Study23 Group Title
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    What does the first derivative determine then?

    • 2 years ago
  13. Study23 Group Title
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    And, how do I determine the values for which this function is positive?

    • 2 years ago
  14. tcarroll010 Group Title
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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.

    • 2 years ago
  15. Study23 Group Title
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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..

    • 2 years ago
  16. Study23 Group Title
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    now*

    • 2 years ago
  17. tcarroll010 Group Title
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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.

    • 2 years ago
  18. Study23 Group Title
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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 \(\ \Huge (\frac{4}{3}, +\infty) ?\)

    • 2 years ago
  19. tcarroll010 Group Title
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    yes, concave up in that interval. The left "(" should be a "["

    • 2 years ago
  20. tcarroll010 Group Title
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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.

    • 2 years ago
  21. Study23 Group Title
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    Okay! I thought it was a ( because the inequality was greater than not greater than or equal to

    • 2 years ago
  22. tcarroll010 Group Title
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    You are understanding this problem well now it seems.

    • 2 years ago
  23. Study23 Group Title
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    :)

    • 2 years ago
  24. tcarroll010 Group Title
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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.

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

    • 2 years ago
  26. tcarroll010 Group Title
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    You are quite welcome! Nice working with you!

    • 2 years ago
  27. Study23 Group Title
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    It was nice working with you, as well :)

    • 2 years ago
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