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neik
Sep24-04, 05:12 PM
Help me plz! :confused:

http://img.photobucket.com/albums/v64/neik7/graph.png

For the function f whose graph is given, arrange the following numbers in increasing order and explain why:


0
f'(2)
f(3) - f(2)
1/2[f(4) - f(2)]

UrbanXrisis
Sep24-04, 05:23 PM
0 is undefined...on an asymptote

out of f'(2), f(3) - f(2) and 1/2[f(4) - f(2)]

increasing order:

1/2[f(4) - f(2)]
f(3) - f(2)
f'(2)

neik
Sep24-04, 07:19 PM
but why? :confused:

vsage
Sep24-04, 07:28 PM
a. 0
b. f'(2)
c. f(3) - f(2)
d. 1/2[f(4) - f(2)]

0 is the smallest and I'll explain why in a second.
f(3) - f(2) can be rewritten as (f(3) - f(2)) / (3-2). This slope is only a tiny fraction smaller than b because if you notice the slope is gradually decreasing as x gets larger. 1/2(f(4)-f(2)) can be rewritten as (f(4)-f(2))/(4-2). This looks like the average slope over that interval which you will notice was slightly less than c (which is in turn slightly less than b)

b, c, and d are positive though because the graph increases from 0-infinity apparently so a is the smallest. Ok I pretty much did all the work for you whoops.