Posted by **Angelina** on Thursday, March 8, 2012 at 3:21pm.

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!

- Math -
**Steve**, Thursday, March 8, 2012 at 5:22pm
Rolle's Theorem states that on the interval (2,3) there will be a point c where f'(c) = the slope of the chord from f(2) to f(3)

So, f(2) = 102

f(3) = 102

so, the chord has slope 0.

so, at some point c in (2,3) f'(c) = 0

f'(t) = 80 - 32t

f'(t) = where t = 80/32 = 2.5

That's what you'd expect, since parabolas are symmetric about their axis.

- Math -
**Angelina**, Thursday, March 8, 2012 at 5:40pm
THNKS :)I really appreciate your help!

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