# Thread: consider the function using Mean Value Theorem

1. ## consider the function using Mean Value Theorem

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?

britt w

2. Originally Posted by williamb
Consider the function f(x) = e^(5x/31) , where the interval is 4 <= x <= 6 ...

... f'(x) = [(e^(20/31)) - (e^(30/31))] / (-2)
what is f'(x) ?

once you find it, set it equal to the average rate of change (the right side of the above equation) , then solve for the value of x in the interval (0,4) that is guaranteed to exist by the MVT.

3. do you mean find the actual NUMBER value of the derivative?
how do i calculate that?
[(e^(20/31)) - (e^(30/31))] / (-2)

when i press "e^x" on my calculator it just returns 1
so do i have to enter it 20/31 e^x for the correct answer?

4. 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?!

5. $\displaystyle f(x) = e^{\frac{5x}{31}}$

$\displaystyle f'(x) = \frac{5}{31}e^{\frac{5x}{31}}$

$\displaystyle \frac{5}{31}e^{\frac{5x}{31}} = \frac{f(6) - f(4)}{6 - 4}$

$\displaystyle \frac{5}{31}e^{\frac{5x}{31}} = \frac{e^{\frac{30}{31}}-e^{\frac{20}{31}}}{2}$

$\displaystyle e^{\frac{5x}{31}} = \frac{31}{10} \left(e^{\frac{30}{31}}-e^{\frac{20}{31}}\right)$

$\displaystyle \frac{5x}{31} = \ln\left[\frac{31}{10} \left(e^{\frac{30}{31}}-e^{\frac{20}{31}}\right)\right]$

$\displaystyle x = \frac{31}{5} \ln\left[\frac{31}{10} \left(e^{\frac{30}{31}}-e^{\frac{20}{31}}\right)\right] \approx 5.027$