Finding the maximum of a function

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The discussion centers on why a function f attains its local maximum at a point r' within the interval (p, q). It is suggested that this occurs because f(x) is less than or equal to f(r') for all x in the interval (p, p+delta). The participants confirm that equality holds at the maximum point. The clarification emphasizes the relationship between the function's values and the definition of a local maximum. Understanding these conditions is crucial for analyzing local maxima in mathematical functions.
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Why does f attain its local maximum at r' in (p,q). Is it because we have f(x)<= f(r') for all x in (p,p+delta)?
 
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ratio said:
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Why does f attain its local maximum at r' in (p,q). Is it because we have f(x)<= f(r') for all x in (p,p+delta)?
We have equality, but, yes.
 
Thanks for you answer^^
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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