Find the derivative using the quotient rule.
Find the critical numbers.
Evaluate v(t) at the critical numbers and at 1 and 4
Lowest value is the minimum, highest the maximum
When I did this I got v'=8t/(1+t^2)^2 so t=0 is a critical number
v(0)=0
V(1)=2
v(4)=64/17
therefore v(4)=64/17 is the maximum
v(0)=0 is the minimum
Look at the graph on your calculator to verify your answers
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.
2 answers
but on the interval 1≤t≤4
v(1) is min, v(4) is max.
v(1) is min, v(4) is max.
2 answers