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October 22nd, 2013, 07:34 AM  #1 
Senior Member Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0  Finding Smallest & Largest Possible Values
woudl this work? Suppose f(0) = 0 and 1 <= f'(x) <= 3 for all x. What is the smallest possible values for f(2)? my solution: What is the largest possible value for f(2)? 
October 22nd, 2013, 08:06 AM  #2 
Senior Member Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0  Re: Finding Smallest & Largest Possible Values
Edited*

October 22nd, 2013, 10:00 AM  #3 
Senior Member Joined: Feb 2013 Posts: 281 Thanks: 0  Re: Finding Smallest & Largest Possible Values
Sorry, but I have to say that your solution is very confusing and wrong. You are right: the starting point is the mean value theorem. But you applied it improperly. You used the formula: f(b)f(a) = f'(c)*(ba). You applied it by b = 2 and a=0, then why did you type 31 at the end?? It should be 20. The third equation is more confusing. Why does the plus sign vanish? Why f'(c) = 2? Here is the solution for the smallest value of f(2). We search for the smallest value for f(2). We know that the function starts off from the origin. You want to keep the function so low as possible, for example you maybe think about a strongly decreasing function. But we have a condition, the f'>=1 at everywhere, so the function can not be descreasing, actually it has to be increasing at least with the slope 1. So intuitively the lowest case is when f(x) = x. For this f(0)=0 and its derivative is f' = 1 which >=1 indeed. If we are right, then the lowest value of f(2) is 2. Let prove this. Let's assume the lowest value is the number L<2. Let apply the mean value theorem to the point (0,2). The theorem states that it must be a point c so: f(2)f(0)=f'(c)*2 that is we have a point c so: L = 2f'(c) therefore we have a point c that: f'(c)=L/2. Now we have a contradiction, if L was less than 2 indeed, then f'(c) would be less than 1, which is not possible, because f'>=1. 

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