Posted by Jake on Saturday, March 3, 2012 at 10:00pm.
If x≤ 4 it should have a max at 4
If x<4 there is no max
since v'(t) = 8t/(1+t^2)^2
v' > 0 for 1<=t<=4
since v is increasing on the interval,
max v is at t=4
min v is at t=1
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