Find the critical point of f(x) given f'(x). With f'(x) being:
Assume f(x) is continuous everywhere.
Now than, my approach was to start by find when f'(x)=0 and when f'(x)=DNE. These are the only X coordinates on f(x) that are critical points. From my calculations I found that critical points are located at X=-7 and X=3. Next I decided to use the first derivative test to determine the behavior of the graph around these critical points and simultaneously hoping to confirm a max or min. The test points I chose to use around -7 were -8 & -6. At X=-8 the sign of f'(x) is negative and therefore slopes down. At X=-6 the sign of f'(x) is positive and therefore slopes upward. Therefore I concluded that X=-7 is a minimum point. I thought about the possibility of X=-7 being a vertical asymptote for f(x) but since it is given that f(x) is continuous I disregarded that possibility. Next I tested X=2 & X=4 around X=3 (the point under consideration). At X=2 I found the sign of f'(x) to be positive and therefore sloping upwards. At X=4 I found the sign of f'(x) to be negative and sloping downwards.
Therefore by my calculations and given the fact that f(x) could not contain asymptotes because it is continuous. I concluded that f(x) had a minimum at X=-7 and a maximum at X=3. If anyone can possibly help me out that would be great. I thought that maybe after typing this up I would have located my mistake and solved the problem correctly myself but I unfortunately am still confused about where I messed up and what parts of this problem I don't seem to understand.
Thank you,
The slope is undefined at x=-7, therefore you can only have a relative maximum at x=3. This is possible without having x=-7 to be a vertical asymptote, it could be a cusp, or a very drastic change in direction, like f(x)=|x| whose slope is undefined at x=0. Being like that, the graph is still continuous. Maybe this will help, maybe not! :P
ReplyDeletethanks man, me and Robert when and talked to him after the solutions were posted. turns out he messed up and typed the solution in wrong. So I did end up getting credit for it. (and also the rectangular prism problem; it turned out to be 2a. which means that for some reason he threw out the square root of 3. not very professional :-)
ReplyDeleteDamn, Dr.C needs to get his act together! haha.
ReplyDelete