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Can f''(x_0) = 0 if f'(x_0) =/= 0?

  1. Dec 6, 2015 #1
    I was wandering, can $$\ \frac { d^{ 2 }y }{ dx^{ 2 } } | _{ x={ x }_{ 0 } }=\quad 0$$ if $$\frac { dy }{ dx }| _{ x={ x }_{ 0 } }\neq \quad 0$$ and if so, what does this translate to geometrically?
     
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  3. Dec 6, 2015 #2

    HallsofIvy

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    Yes, of course. That is the same as saying that a function can have the value "0" at a point where its derivative is not 0. And it is very easy to give an example of that- suppose y= x- 1. That is a straight line such that y(1)= 1-1= 0 but its derivative is the constant 1 which is never 0. To relate that to your question about first and second derivatives, take an "anti-derivative". An anti-derivative of x- 1 is y= (1/2)x^2- x+ 1. Now, y'= x- 1 which is 0 at x= 1 while the second derivative is 1.

    An nth derivative tells how fast the n-1 derivative is changing and is NOT related to the actual value of that n-1 derivative.
     
  4. Dec 6, 2015 #3

    Samy_A

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    ##y=x-x_0##
     
  5. Dec 6, 2015 #4
    Thanks. But I was hoping for a non linear function. Or in other words, I am wandering what this means geometrically. I know that if a point is an inflection point then its second derivative is 0 but the converse doesn't necessarily hold. In other words, what does f''(x_0) = 0 tell us about that point other than it being a likely inflection point.

    Thanks :)
     
  6. Dec 6, 2015 #5

    FactChecker

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    This means that the slope of the curve is not changing at that point
    This means the slope is not zero at the point. So @Samy_A 's example of a sloped straight line satisfies both conditions at every point.
     
  7. Dec 6, 2015 #6

    FactChecker

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    Nothing. That is exactly what it means. sin(x) has an inflection point at every x= n * pi. Those are all inflection points with non-zero slopes.
     
  8. Dec 6, 2015 #7

    WWGD

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    Actually, take a degree-2 or higher polynomial other than ##x^n##, i.e. ##a_nx^n+...+a_1x+a_0 ; a_j## not all 0 ## n>j\geq 0##, I would say the probability of having both f'(x) and f''(x)=0 is 0 under "reasonable" choices of pdf..
     
  9. Dec 7, 2015 #8
    I see. Thanks all!
     
  10. Dec 22, 2015 #9

    LCKurtz

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    The second derivative geometrically identifies concavity. f'' > 0 or f''<0 on an interval identifies concave up or down, respectively. In a case where f'' is continuous, the only way for concavity to switch is for f'' to pass through zero. So f'' = 0 at a point might be a case where f'' is changing sign, indicating an inflection point (change of concavity). Of course, f'' might not change sign nearby so it might not be an inflection point. I wouldn't say such a point is a "likely" inflection point but a "possible" inflection point.
     
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