February 15th, 2019, 06:36 AM  #11  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
$\text {IF } f(x) \text { is continuous on } [u, v], \text { differentiable on } (u, v),$ $u \le x < w \implies f'(x) > 0, \text { and } w < x \le v \implies f'(x) < 0,$ then you say that it may be false that f(x) is maximized on the integers at either the floor or else the ceiling of w. That is essentially the situation on this problem except we have an open interval. I do not see that your counterexample is relevant to this problem. I can see that things become more complex if there are multiple points in an interval where the function's derivative is zero, but, for a differentiable function on an interval that includes at least one integer, how can the maximum on the integers not be near an endpoint or a critical point? I don't understand "completely invalid." Perhaps "over simplified" is intended. Or, perhaps, my intuition is playing me completely false. Last edited by JeffM1; February 15th, 2019 at 06:38 AM.  
February 15th, 2019, 08:00 AM  #12 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 
Please remain on topic.

February 15th, 2019, 01:30 PM  #13  
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 723  Quote:
 
February 15th, 2019, 02:20 PM  #14 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
For the third image : Calculate area $\displaystyle S_1$ of curve y1 bounded by xaxis . Calculate the area $\displaystyle S_2$ of line y2 bounded by xaxis and y1 . The enclosed area between y1 and y2 is $\displaystyle S_1 S_2$ . Last edited by idontknow; February 15th, 2019 at 02:27 PM. 
February 16th, 2019, 05:37 AM  #15 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
You must find the intersection points of y1 and y2 . Just equal them y1=y2 and solve for x. 
February 16th, 2019, 06:15 AM  #16 
Newbie Joined: Feb 2019 From: Scotland Posts: 4 Thanks: 0  
February 16th, 2019, 09:00 AM  #17 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
Equation y1=y2 has solutions 4 and 3 . $\displaystyle S=\int_{4}^{3}( y_1 y_ 2 )dx$ . 

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