# Thread: Mean Value Theorem

1. ## Mean Value Theorem

my problem I am working on is f(x)= x+(54/x); [6,9]

First I took the first derivative and found that it was f ' (x)= 1-(54/x^2)

I then found f(6) = 15 and f(9) = 15

If I plug those into the formula f(b)-f(a) I get 0 this just doesn't make
b-a 3

This just doesn't make sense!! Could someone help me?

2. What doesn't make sense?

$\displaystyle f'(c) = \frac{f(b)- f(a)}{b-a} \: \: c \in (a,b)$

Now, you found that f'(c) = 0, i.e. there exists some point on your interval such that its derivative is equal to 0. What does it exactly mean for your function to have its derivative equal to 0 at c?

Hint: Rolle's Theorem, a special case of the mean value theorem.

3. Originally Posted by kelleannmoore@yahoo.com
my problem I am working on is f(x)= x+(54/x); [6,9]

First I took the first derivative and found that it was f ' (x)= 1-(54/x^2)

I then found f(6) = 15 and f(9) = 15

If I plug those into the formula f(b)-f(a) I get 0 this just doesn't make
b-a 3

This just doesn't make sense!! Could someone help me?
The Mean Value Theorem states that if $\displaystyle f(x)$ is continous on some interval $\displaystyle D=[a,b]$ and differentiable on $\displaystyle (a,b)$ then

$\displaystyle \exists{c}\in{D}\backepsilon~f'(c)=\frac{f(b)-f(a)}{b-a}$

So we see that we have a special case of the Mean Value Theroem called Rolle's Theorem that states, that if our function meets the same conditions as before and that $\displaystyle f(a)=f(b)$

Then $\displaystyle \exists{c}\in{D}\backepsilon~f'(c)=0$