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,
britt w
k this is where i'm at..
so i take my first derivative (slope of secant line), which is on the RHS
and i equate it to the slope of my tangent line, which is on the LHS
(e^(5x/31)*5/31) = (e^(20/31)-e^(30/31))/-2
i need to solve for x, i THINK i may have gotten it.. so far.. who knows:
(e^(5x/31)*5/31) = -0.7257000/-2
(e^(5x/31)*5/31) = 0.362850029
e^(5x/31) = 2.249670184
take the ln of each to bring the exponent down, and solve
5x/31 = ln(2.249670184)
x = [31 * ln(2.249670184)]/5
so now i...
plug x back into (e^(5x/31)*5/31) = (e^(20/31)-e^(30/31))/-2 to get my f'(x)??
(e^(5([31 * ln(2.249670184)]/5)/31)*5/31) = (e^(20/31)-e^(30/31))/-2
in the exponent i suppose we could simplify to get
(e^(ln(2.249670184))*5/31) = (e^(20/31)-e^(30/31))/-2
and then just transfer over to one side for our f'(x)???
f'(x) = [(e^(20/31)-e^(30/31))/-2 ] - [(e^(ln(2.249670184))*5/31)]
if this is the case and in some nuts way i'm correct, how do we find the c value?!