find the value ?

Question

find the value of c guaranteed by the mean value theorem for y=cosx on the interval [0,pi/2]

mean value thm says that f(x) is continuous on that closed interval, and differentiable on the open interval.
also, f(b)-f(a)/(b-a)=f ' (c)
i did the last hting out, and got f '(c) to equal -.6366, is this right? i dont know what im doing!

you can do that even though sinx is negative? how do i figure out the value of x whicih makes -sinx equal -.63...whatever it was?

Answer 1

Yeah, thats a reasonable start. The mean value theorem says that there must be a value of c in [0,pi/2] with (cos(pi/2) - cos(0))/(pi/2 - 0) = f'(c).
Working that out gives f'(c) = -1/(pi/2) = -2/pi = -0.6366 as you said.
But you want the value of c.
f'(c) = -sin c, so you want sin c = 0.6366, which gives c = 0.6901.

Responding to your question..
if -sin x = -0.6366, then sin x = 0.6366, right? Then just do sin^-1 on a calculator.

Answer 2

f' = -sin x

[ f(0) - f(pi/2)]/pi/2 = -2/pi and this is about -.6