Consider the function f(x) = e^(5x/31) , where the interval is 4 <= x <= 6
(a) We know that the Mean Value Theorem can be applied to (f) on the interval [4,6] because (f) is ______ on [4,6] and _____ on (4,6).
Note. You are to enter two mathematical terms as answers.
(b) If we now apply the Mean Value Theorem to (f) on the interval [4,6], we can conclude that there is at least one number (c) in the interval (4,6) such that f'(c) =_______.
(c) Find all possible value(s) for such (c)
Okay so i know that the MVT says that a continuous function on [a,b] an differentiable on (a,b) where a <b then there exists a c value in (a,b) such that f'(c) = [f(b) - f(a)] / (b-a)
I entered my a and bvalues into teh derivative equation as given in the MVT,
f'(x) = [(e^(5*4/31)) - (e^(5*6/31))] / (4-6)
--> f'(x) = [(e^(20/31)) - (e^(30/31))] / (-2)
but i just don't know where to go from here..
do i equate it to zero?
this is a new concept for us, please help
thanks in advance,