Posted by Jake on Saturday, March 3, 2012 at 10:00pm.
An object moves in a straight line. Its velocity, in m/s, at time t is v(t)=4t^2/(1+t^2),t≥0.
Determine the maximum and minimum velocities on the interval 1≤x≤4.
The answer says min. velocity is 0m/s but has no maximum?

Calculus  Susan, Sunday, March 4, 2012 at 8:38am
If x≤ 4 it should have a max at 4
If x<4 there is no max

Calculus  Steve, Sunday, March 4, 2012 at 12:05pm
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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