MHB 2.7.2 AP Calculus Exam .... distance traveled estimate

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The discussion centers on estimating the distance traveled in a 2.7.2 AP Calculus Exam problem. A participant calculated the distance by counting squares on a graph, determining that there are approximately 21 squares. Each square represents 10 feet, leading to a total distance estimate of 210 feet. The method of counting squares is confirmed as a valid approach for this type of problem. This straightforward calculation illustrates a practical application of graphical interpretation in calculus.
karush
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ok all I did was count squares so about 21
 
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karush said:
ok all I did was count squares so about 21

every square represents 10 ft, so 21 times 10 ft ?
 
210:cool:
 
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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