# Mean Value Theorem

• Mar 31st 2010, 07:18 PM
sderosa518
Mean Value Theorem
f(x)= square root (X) - 2x

Stupid error I made and I lost myself. Can someone assist please?

Once I got the dirivative what do I need to do next?

(1/2x)^-1/2
------------ = 1/2
1/2
• Mar 31st 2010, 07:23 PM
Prove It
Quote:

Originally Posted by sderosa518
f(x)= square root (X) - 2x

Stupid error I made and I lost myself. Can someone assist please?

Once I got the dirivative what do I need to do next?

(1/2x)^-1/2
------------ = 1/2
1/2

The mean value theorem states that

$f'(c) = \frac{f(b) - f(a)}{b - a}$ for some $c \in [a, b]$.

The question is, what are you trying to do with the MVT? Have you been given two points and asked what is the point in between them where the derivative is the same as the slope of the secant?
• Mar 31st 2010, 07:27 PM
sderosa518
Use the mean value theorem

f(x)= square root (x) - 2x [0,4]

Sorry about the wrond input. I know that I am going through the right path, but I got lost.
• Mar 31st 2010, 07:37 PM
Prove It
$f(x) = \sqrt{x} - 2x$.

You need to find $f(0), f(4), f'(x)$.

$f(0) = \sqrt{0} - 2(0)$

$= 0$.

$f(4) = \sqrt{4} - 2(4)$

$= 2 - 8$

$= -6$.

$f'(x) = \frac{1}{2\sqrt{x}} - 2$.

So by the Mean Value Theorem:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

$\frac{1}{2\sqrt{c}} - 2 = \frac{-6 - 0}{4 - 0}$

$\frac{1}{2\sqrt{c}} - 2 = \frac{-6}{4}$

$\frac{1}{2\sqrt{c}} - 2 = -\frac{3}{2}$

$\frac{1}{2\sqrt{c}} = \frac{1}{2}$

$2\sqrt{c} = 2$

$\sqrt{c} = 1$

$c = 1$.

So by the Mean Value Theorem:

$f'(1) = \frac{f(4) - f(0)}{4 - 0}$.