posted by Angelina .
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)?
& 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!
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.
THNKS :)I really appreciate your help!