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Homework Help: Determine the values of a and b that have inflection points

  1. Dec 12, 2017 #1
    <Moderator's note: Moved from a technical forum and thus no template.>

    The title isnt complete this is what I meant to say:
    Determine the values of aa and bb where the function has inflection points (x2+ax+b)(ex)

    I made the second derivative
    $$f''(x) = 2 e^x + 2 a e^x + b e^x + 4 e^x x + a e^x x + e^x x^2$$

    and isolate the x:

    $$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)$$

    And a guy told me this:
    ---------------------------------------
    Now, you have inflection points if $$√a2−4b+8$$ is real. Or $$a2−4b+8≥0$$.
    However, If $$a2−4b+8=0$$ then f′′(x) does not change sign when x crosses the root, $$a2−4b+8>0$$.
    ----------------------------------------
    And now Im lost because i've trying to figure out why if the root is 0 then f''(x) doesnt change sign? What's left in the fraction doesn't matter?
     
    Last edited by a moderator: Dec 12, 2017
  2. jcsd
  3. Dec 12, 2017 #2

    FactChecker

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    f'' is a second order polynomial. If it has two zeros then it crosses the line f'' = 0 at one zero, changes sign, turns around, comes back to the other zero, and changes sign again. If it only has one zero, then it just barely touches the f''=0 line, turns around without crossing it and stays on one side without changing sign. But the only way f'' can only have one zero is if ±√ is only one value. But that can only happen if it is ±0.
     
  4. Dec 12, 2017 #3
    I still don't understand what do you mean by two zeros? The 0 of the square root and the f''=0 ? Im not able to visualize that in my head
     
  5. Dec 12, 2017 #4

    SammyS

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    What are the intermediate steps between the above two equations ?
     
  6. Dec 12, 2017 #5
    The original function:

    $$f(x) = (x^2+ax+b)(e^x)$$

    Then:

    $$f ' (x)= (x^2 +ax + b)(e^x)+(e^x)(2x+a)$$
    $$f ' (x)= e^x(x^2+2x+ax+a+b)$$

    Then :

    $$f '' (x)= (x^2+2x+ax+a+b)(e^x) + e^x(2x+2+a)$$
    $$f '' (x)= e^x(x^2+4x+ax+2a+b+2)$$

    And isolated the x with the formula for quadratic polynomial:

    where $$a= 1, b=4 + a, c= 2a+b+2$$
    Which leads to:

    $$x=\frac{1}{2}(±√(a2−4b+8)−a−4)$$

    I already finished my homework but still need to know why $$a^2 - 4b + 8$$ cannot be = 0
     
    Last edited: Dec 12, 2017
  7. Dec 12, 2017 #6

    FactChecker

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    ±
    This figure has two examples. The green parabola has two zeros at points A and B. The red parabola has 1 zero (actually a double zero at the same point) at point C. When the quadratic formula is used, the green line will give x values for zeros at A and B where ±√ is about ±1.5. The quadratic formula for the red line will give x=2 for a zero and ±√0. (In the red-line case, that is called a "double zero" at x=2.)
    temp.png
     
    Last edited: Dec 12, 2017
  8. Dec 12, 2017 #7

    SammyS

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    What condition on ##\ f '' (x) \ ## is necessary, for there to be an inflection point at ## x ## ?
    You are using ## a ## and ## b ## each for two different things.

    I suggest using upper case for the coefficients of a general quadratic function.
     
  9. Dec 12, 2017 #8

    FactChecker

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    He is not using the standard a,b,c of the quadratic formula. If A, B, C are the quadratic formula coefficients, then A=1, B=4+a, C=2+2a+b. I think that his values for the zeros is correct.
     
  10. Dec 12, 2017 #9

    SammyS

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    I realize that.

    Does OP?
     
  11. Dec 12, 2017 #10
    Sorry but I don't understand why It works if the square root is 0, what's left in the x doesn't matter?: $$\frac{-4-a}{2}$$

    The whole x shouldn't be equal to 0?

    Yes
     
  12. Dec 12, 2017 #11

    SammyS

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    @Shai ,

    What is an inflection point?
     
  13. Dec 12, 2017 #12
    A point where the concavity of a continuous function change
     
  14. Dec 12, 2017 #13

    SammyS

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    Yes.

    So the second derivative changes sign in going from one side of the inflection point to the other.

    Now consider FC's graphs in Post # 6 .
     
  15. Dec 12, 2017 #14
    Yes but why the sign doesn't change if the square root is equal to 0 but works with any other result of the square root?

    $$x=\frac{1}{2}(±√(a2−4b+8)−a−4)$$

    $$x=\frac{1}{2}(±√(0)-a-4)$$
     
  16. Dec 12, 2017 #15

    SammyS

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    Because, in that case the quadratic function has only one zero (x-intercept).
    If the quadratic function has only one zero, then it doesn't change sign. See the red graph in post #6. Study @FactChecker 's explanation again and again.
     
  17. Dec 13, 2017 #16

    FactChecker

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    That part certainly does matter. If ±√ = ±0, It tells you where the double zero is located. Otherwise, there are two different zeros and it tells you where the midpoint of the two zeros are.

    But it doesn't change the important aspect that you are concerned about -- if the sign of f'' changes. That is just determined by ±√.
     
  18. Dec 13, 2017 #17
    So because $$f ''(x) $$ is a quadratic polynomial the x must have two solutions (greater than 0) in order to have inflection points?
     
  19. Dec 13, 2017 #18

    FactChecker

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    Exactly. -- The √ must be > 0 so that ±√ forces two different zeros.
     
  20. Dec 13, 2017 #19
    At the end it wasn't as hard as I thought, thank you very much and sorry for taking that long to understand.
     
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