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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