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Math Help - tricky tangent qroblem :)

  1. #1
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    tricky tangent qroblem :)

    Suppose g(x) has a second derivative g''(x) at all points of an interval I. Prove that if the tangent to the curve y=g(x) at a in I meets the curve again at some other point b>a in I then g''(x)=0 somewhere between a and b...

    I seriously dont know where to begin
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Khonics89 View Post
    Suppose g(x) has a second derivative g''(x) at all points of an interval I. Prove that if the tangent to the curve y=g(x) at a in I meets the curve again at some other point b>a in I then g''(x)=0 somewhere between a and b...

    I seriously dont know where to begin

    The phrase 'meets the curve again' refers to inflection. So, I would start by understanding that wherever the tangent line to a graph crosses the graph, f''(x)=0. There is a proof of this fact here:

    Second derivative test - Wikipedia, the free encyclopedia
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  3. #3
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    Hey von-

    I'm sorry but I still dont understand what your trying to tell me

    I dont get the link-

    can you explain wat exact is the question asking me.
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  4. #4
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    Quote Originally Posted by Khonics89 View Post
    Suppose g(x) has a second derivative g''(x) at all points of an interval I. Prove that if the tangent to the curve y=g(x) at a in I meets the curve again at some other point b>a in I then g''(x)=0 somewhere between a and b...

    I seriously dont know where to begin
    first note that \frac{g(b)-g(a)}{b-a}=g'(a), because they are both equal to the slope of the tangent line to the curve at x=a. call this (1). define f(x)=g(x)-\frac{g(b)-g(a)}{b-a}(x-a)-g(a), \ x \in I.

    see that f(a)=f(b). thus by Rolle's theorem there exists a<c<b such that g'(c)-\frac{g(b)-g(a)}{b-a}=f'(c)=0, i.e. g'(c)=\frac{g(b)-g(a)}{b-a}. therefore g'(a)=g'(c), by (1). again, applying Rolle's

    theorem, gives us that g''(d)=0, for some a<d<c, and we're done.
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