The height of a ball t seconds after it is thrown upward from a height of 6 feet and with an initial velocity of 80 feet per second is f (t) = -16t^2 + 80t + 6.

According to Rolle's Theorem, what must be the velocity at some time in the interval (2, 3)?
ft/sec?

& Find that time t?

I believe i need to take the 1st derivative which is -32t+80 equal to zero so t=-80/32? and for ft/sec plug in 2 for t? Anyhow i get wrong answers so need help!

You are correct on the time, but it needs to be simplified if not put into decimal form. Now you need to plug in 2 into the orignal problem and get a resultant, then plug in three to the original problem as well. If they are the same resultant then they follow rolles theorem. There are some rules to this theorem like is it a continuous function, is it differentialable? Keep those things in mind because if you go on to Calculus 2, you will see these again in integratable series.

To find the velocity at some time in the interval (2, 3) using Rolle's Theorem, we need to find the derivative of the height function f(t).

The derivative of f(t) is f'(t) = -32t + 80.

According to Rolle's Theorem, if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that f'(c) = 0.

In this case, we are interested in finding the velocity, which corresponds to the derivative f'(t), so we want to find a value c such that f'(c) = 0 in the interval (2, 3).

To find the velocity at some time in the interval (2, 3), we need to find a value of t in the open interval (2, 3) where f'(t) = 0.

Let's solve the equation f'(t) = 0:

-32t + 80 = 0

Adding 32t to both sides:
80 = 32t

Dividing both sides by 32:
t = 80/32

Simplifying:
t = 2.5

So, according to Rolle's Theorem, there must be a time t = 2.5 in the interval (2, 3) where the velocity is 0 ft/sec.

To find the velocity at some time in the interval (2, 3) using Rolle's theorem, we first need to find the derivative of the height function f(t).

The derivative of f(t) = -16t^2 + 80t + 6 represents the velocity function v(t) since velocity is the rate of change of position over time.

Taking the derivative of f(t) with respect to t, we get:

f'(t) = d/dt (-16t^2 + 80t + 6)
= -32t + 80

Now, we can find the time when the velocity is zero. This occurs when f'(t) = 0. So, we solve the equation:

-32t + 80 = 0

Simplifying the equation, we have:

-32t = -80
t = -80 / -32
t = 2.5 seconds

This means that there must be a time t = 2.5 seconds within the interval (2, 3) at which the velocity is zero, according to Rolle's theorem.

To find the velocity at t = 2.5 seconds, we substitute this value back into the velocity function:

v(2.5) = -32(2.5) + 80
= -80 + 80
= 0 ft/sec

Therefore, the velocity at time t = 2.5 seconds in the interval (2, 3) is 0 ft/sec.