Prove f(0)=f'(0)=f"(0)=0 in (-1,1) interval w/ Positive M

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SUMMARY

The discussion centers on proving that for a function f:(-1,1)→R, which is three times differentiable and satisfies the condition ⎮f(x)⎮ ≤ M⎮x⎮³ for a positive constant M, it follows that f(0)=f'(0)=f"(0)=0. The proof begins by establishing that ⎮f(0)⎮≤M(0)=0, leading to f(0)=0. Further, by analyzing the limit of f(x)/x as x approaches 0, it is concluded that f'(0)=0. The next step involves determining f"(0) through the limit of the derivative, requiring consideration of the behavior of f'(x) around zero.

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Homework Statement


Suppose f:(-1,1)→R is three times differentiable on the interval. Assume there is a positive value M so that ⎮f(x)⎮ ≤ M⎮x⎮³ for all x in (-1,1). Prove that f(0)=f'(0)=f"(0)=0.


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The Attempt at a Solution


My professor started us off,
⎮f(0)⎮≤M(0)=0; f(0)=0
f'(0)=lim as x→0 [(f(x)-f(0))/x-0 = lim as x→0 [f(x)/x].
I know that ⎮f(x)/x⎮≤ 1/⎮x⎮(M⎮x⎮³
≤ Mx²
Which means that f'(0) = 0
I also know that the next step is to find f"(0) = lim as x→0 [(f'(x)-f'(0))/x-0].
I need to know if f'(x) = 1/x?
 
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f'(x) is not 1/x. Think about what f(x) is for x > 0 and for x < 0. Then think about what f'(x) should be for x > 0 and for x < 0. This should help you continue the problem. When you have to show f''(0) = 0, again think about what f''(x) should be for x > 0 and for x < 0.
 

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