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

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In summary, the conversation is about proving that if a function f is three times differentiable on the interval (-1,1) and satisfies the condition ⎮f(x)⎮ ≤ M⎮x⎮³, then f(0)=f'(0)=f"(0)=0. The conversation discusses finding f'(0) and f"(0) using the given conditions and the properties of the function f. It is mentioned that f'(x) is not equal to 1/x and that understanding the behavior of f(x) and f'(x) for different values of x can help in proving the statement.
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Janez25
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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.


Homework Equations





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.
 

1. What does it mean to prove f(0)=f'(0)=f"(0)=0 in the interval (-1,1)?

Proving that f(0)=f'(0)=f"(0)=0 in the interval (-1,1) means that the function f has a zero value, first derivative, and second derivative at the point x=0 within the interval from -1 to 1. This indicates that the function is continuous, differentiable, and has a point of inflection at x=0 within the given interval.

2. How do you prove f(0)=f'(0)=f"(0)=0 in the interval (-1,1)?

To prove f(0)=f'(0)=f"(0)=0 in the interval (-1,1), you must show that the function f satisfies the three conditions simultaneously. This can be done by evaluating f(0), f'(0), and f"(0) and showing that they all equal zero within the given interval. Additionally, you may also use the definition of continuity, differentiability, and point of inflection to prove these conditions.

3. Why is it important to prove f(0)=f'(0)=f"(0)=0 in the interval (-1,1)?

Proving that f(0)=f'(0)=f"(0)=0 in the interval (-1,1) is important because it provides information about the behavior of the function at a specific point within a given interval. It also indicates that the function is smooth and has a point of inflection at x=0 within the interval, which can be useful in analyzing the behavior of the function.

4. Can f(0)=f'(0)=f"(0)=0 in the interval (-1,1) if the function is not continuous, differentiable, or has a point of inflection at x=0?

No, f(0)=f'(0)=f"(0)=0 in the interval (-1,1) can only occur if the function is continuous, differentiable, and has a point of inflection at x=0 within the given interval. If any of these conditions are not satisfied, then the function cannot have a zero value, first derivative, and second derivative at x=0 simultaneously.

5. What is the significance of having a positive M in the statement "Prove f(0)=f'(0)=f"(0)=0 in (-1,1) interval w/ Positive M"?

The positive M in this statement indicates that the function f has a positive slope at x=0 within the interval (-1,1). This means that the function is increasing at x=0 and is a useful piece of information when analyzing the behavior of the function within the given interval. It also helps to further narrow down the behavior of the function at x=0 and provides a clearer understanding of its properties within the interval.

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