# Differential calculus question (Mean value theorem)

1. May 28, 2005

### janiexo

You are given the following information about the function f(x):
i) There is an x-value x* approximately equal to 0.8 such that f(x*)=0
ii) f(0.7) = C is negative
iii) m1 < f'(x) < m2 for 0.7 < x < 0.9 where m1 and m2 are positive constants

Apple the Mean Value Theorem to f(x) on the interval [0.7,x*] to find upper and lower bounds (in terms of m1, m2 and C) for x*

I've been really struggling to understand the MVT and have been at this question for a while... i just can't seen to work it out though and end up going round in circles I have a test coming up so i really want to try to get my head around the MVT.

2. May 28, 2005

### Andrew Mason

By the Mean Value Theorem, there exists some w between .7 and x* such that:

$$f'(w) = \frac{f(x*) - f(.7)}{x* - .7} = \frac{0 - C}{x* - .7}$$

Since m1 < f'(w) < m2,

$$m1 < \frac{- C}{x* - .7} < m2$$

You should be able to work out the upper and lower bounds of x* from that.

AM

3. May 29, 2005

### janiexo

Great, that's what I did, only I keep getting the answer 0.7-C/m1 < x* < 0.7 - C/m2 but the actual answer is 0.7-C/m1 > x* > 0.7 - C/m2. I'm probably making some stupid mistake, but this is what I did:

m1 < -C/(x*-0.7) < m2
1/m1 > (x*-0.7)/-C > m2 (i changed the signs because i took the reciprocol)
-C/m1 < (x*-0.7) < -C/m2 (i changed the signs because * by a negative number)
0.7-C/m1 < x* < 0.7-C/m2

Can you tell me where I fudged up? It's probably really obvious but I can't see it :(

4. May 29, 2005

### shmoe

-C is positive

5. May 29, 2005

### Andrew Mason

Take each inequality separately and do the algebra to separate x*.

$$m1 < \frac{-C}{(x*-0.7)}$$

$$x* - .7 < \frac{-C}{m1}$$

$$x* < .7 - \frac{C}{m1}$$

Similarly:

$$x* > .7 - \frac{C}{m2}$$

so:
$$.7 - \frac{C}{m1} > x* > .7 - \frac{C}{m2}$$

AM

6. May 30, 2005

Thank you :)