intermediate value theorem

**arbolis** · May 14th 2008, 04:53 PM

Check out Intermediate value theorem - Wikipedia, the free encyclopedia.

**szpengchao** · May 14th 2008, 04:55 PM

i m actually not meaning the first part...

**arbolis** · May 14th 2008, 05:04 PM

Ok. Next time be more precise then

. Take a look at Mean value theorem - Wikipedia, the free encyclopedia. You must prove this theorem... called the mean value theorem. (or something very close to it).

**szpengchao** · May 14th 2008, 05:10 PM

is there any connection between mean value theorem and the thing we asked to prove ?

i mean the f(b') - f(a') / ... = k

**arbolis** · May 14th 2008, 05:19 PM

is there any connection between mean value theorem and the thing we asked to prove ?

Yes. The mean value theorem states that there exist a $c$ in $(a,b)$ such as $\frac{f(b)-f(a)}{b-a}=f'(c)$ . In your example, $g$ and $h$ are defined and continuous on $(a,b)$ , you must find a subinterval of $(a,b)$ in which the mean value theorem can be applied.

**szpengchao** · May 14th 2008, 06:38 PM

i still cant see any obvious connection...
anyone who can do it for me please?

**szpengchao** · May 14th 2008, 06:42 PM

be careful. hav u seen that f'(a) < k < f'(b)

that is f', not f

**ThePerfectHacker** · May 14th 2008, 07:30 PM

The intermediate value theorem for derivatives states that if $f:I\mapsto \mathbb{R}$ is a differenciable function then $f'$ has the intermediate value property on $I$ , an open interval. Meaning if $a,b\in I$ with $a<b$ and $c$ lies (strictly) between $f'(a)$ and $f'(b)$ there exists $x_0\in (a,b)$ so that $f'(x_0) = c$ .

Without lose of generality we will assume that $f'(a) < c < f'(b)$ . Now define the function $g:[a,b]\mapsto \mathbb{R}$ as $g(x) = f(x) - cx$ . Since $g$ is continous on $[a,b]$ it assumes a minimum value. We will now show that $g$ does not take a minimum value at $a$ . Note that $g'(a) = f'(a) - c < 0$ . But since $g'(a) = \lim_{x\to a^+}\tfrac{g(x)-g(a)}{x-a} < 0$ it follows that $g(x) - g(a)<0$ for $a<x$ sufficiently close to $x$ . Thus, there exists $y\in (a,b)$ so that $g(y) - g(a) < 0$ , which is means $g(y) < g(a)$ , in particular $a$ is not a minimum for $g$ . Using a similar argument we can show $b$ is not a minimum for $g$ . Thus, $g$ has a minimum on $(a,b)$ and so there is a point $x_0 \in (a,b)$ so that $g'(x_0) = 0\implies f'(x_0) = c$ .

Thread: intermediate value theorem

LinkBack

Thread Tools

Display

intermediate value theorem

oh..

dotdot

cant see

be careful

Similar Math Help Forum Discussions

Intermediate Value Theorem

Intermediate Value Theorem

Intermediate Value Theorem

Intermediate Value Theorem

intermediate value theorem/rolle's theorem

Search Tags