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Thread: 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, $\displaystyle f''(x)=0$. 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 $\displaystyle \frac{g(b)-g(a)}{b-a}=g'(a),$ because they are both equal to the slope of the tangent line to the curve at $\displaystyle x=a.$ call this (1). define $\displaystyle f(x)=g(x)-\frac{g(b)-g(a)}{b-a}(x-a)-g(a), \ x \in I.$

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

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