Assume that f is non-negative on (0,1)

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SUMMARY

The discussion centers on a mathematical problem involving a non-negative function f defined on the interval (0,1) with its third derivative existing in that range. The user seeks to demonstrate that if f(x) equals zero at two distinct points within this interval, there exists a point c in (0,1) where the third derivative f'''(c) equals zero. The application of Rolle's Theorem is highlighted, indicating that between the points where f vanishes, there exists at least one point p where the first derivative f'(p) is zero, leading to further implications about the behavior of the function.

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  • Understanding of Rolle's Theorem in calculus
  • Knowledge of derivatives and their significance
  • Familiarity with the properties of non-negative functions
  • Basic concepts of local minima and maxima
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  • Explore the conditions under which a function's higher derivatives can be zero
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mansi
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Hello...I need help and I know this is a very simple problem...I don't know why I'm getting stuck( Maybe because it's past midnight here:frown: )

Assume that f is non-negative on (0,1) and the third derivative of f exists on (0,1).If f(x)=0 for at least 2 values of x in (0,1), show that there exists "c" in (0,1) such that the third derivative of f at c is 0.

this is what I've done...let a,b be the points where f vanishes. that is f(a)=f(b)=0
so we can use Rolle's theorem and get a point p such that f^(p)=0.

What next??
 
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draw a picture: hint it is non-negative and touches the x-axis twice (necessarily at local minima)
 

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