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?

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

Well, well, well, it seems like we have a sneaky little situation with no maximum velocity! How peculiar! Perhaps this object doesn't believe in limits and just wants to keep on going and going, like the Energizer Bunny!

But fear not, my friend, we can still determine the minimum velocity. Now, to find the minimum velocity on the interval from 1 to 4, we need to find the lowest point on the graph of the velocity function. That's where the clownish trick comes into play!

Now, if you analyze the function v(t) = 4t^2/(1+t^2), you'll notice that it approaches 0 as t approaches both positive and negative infinity. However, we need to find the minimum velocity on the interval from 1 to 4.

So, let's first check the endpoints of the interval. When t = 1, we have v(1) = 4/(1+1) = 2 m/s. And when t = 4, we have v(4) = 4*16/(1+16) = 64/17 m/s.

Now, if we take a look at the graph of the velocity function, we can see that it has a point where it reaches the value of 0. This happens at t = 0. Ah, the sneaky clown at it again!

But here's the catch: the interval we're considering is from t = 1 to t = 4. Since 0 is not within this interval, we cannot consider it as a minimum velocity on this interval. Thus, the minimum velocity on the interval 1 ≤ t ≤ 4 is indeed 2 m/s.

So, my friend, we may not have a maximum velocity on this interval, but at least we have found the minimum velocity! Keep those wheels turning, and remember, sometimes the clownish tricks of math can leave us with a good chuckle!

To determine the maximum and minimum velocities on the interval 1≤t≤4, we first need to find the critical points of the velocity function v(t). The critical points occur where the derivative of v(t) equals zero or is undefined.

Let's find the derivative of v(t):
v'(t) = d/dt (4t^2/(1+t^2))

To simplify the derivative, we can use the quotient rule, which states that if f(t) = g(t)/h(t), then f'(t) = (g'(t)h(t) - g(t)h'(t))/[h(t)]^2.

Applying the quotient rule:
v'(t) = [(2)(4t)(1+t^2) - (4t^2)(2t)] / [(1+t^2)]^2
= [8t(1+t^2) - 8t^3) / [(1+t^2)]^2
= [8t + 8t^3 - 8t^3] / [(1+t^2)]^2
= 8t / [(1+t^2)]^2

To find the critical points, we set the derivative v'(t) equal to zero:
8t / [(1+t^2)]^2 = 0

Since the numerator is the only part that can be zero, we have:
8t = 0

Therefore, the only critical point occurs at t = 0.

Next, we need to evaluate the velocity function at the endpoints of the interval, t = 1 and t = 4.

v(1) = 4(1)^2 / (1 + (1)^2) = 4/2 = 2 m/s
v(4) = 4(4)^2 / (1 + (4)^2) = 4(16) / 17 = 64/17 ≈ 3.765 m/s

Now, we can determine the minimum and maximum velocities:

Minimum velocity: The minimum occurs either at the critical point (t = 0) or at one of the endpoints. Comparing the values, we find that the minimum velocity is 0 m/s, which occurs at t = 0.

Maximum velocity: Since the function has no critical points other than t = 0, we can say that the maximum velocity occurs at the endpoint t = 4, which is approximately 3.765 m/s.

Therefore, the minimum velocity is 0 m/s, and the maximum velocity on the interval 1≤t≤4 is approximately 3.765 m/s.