# intermediate value theorem

• May 14th 2008, 04:27 PM
szpengchao
intermediate value theorem
• May 14th 2008, 04:53 PM
arbolis
• May 14th 2008, 04:55 PM
szpengchao
oh..
i m actually not meaning the first part...
• May 14th 2008, 05:04 PM
arbolis
Ok. Next time be more precise then (Wink). 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).
• May 14th 2008, 05:10 PM
szpengchao
dotdot
is there any connection between mean value theorem and the thing we asked to prove ?

i mean the f(b') - f(a') / ... = k
• May 14th 2008, 05:19 PM
arbolis
Quote:

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 $\displaystyle c$ in $\displaystyle (a,b)$ such as $\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c)$. In your example, $\displaystyle g$ and $\displaystyle h$ are defined and continuous on $\displaystyle (a,b)$, you must find a subinterval of $\displaystyle (a,b)$ in which the mean value theorem can be applied.
• May 14th 2008, 06:38 PM
szpengchao
cant see
i still cant see any obvious connection...
anyone who can do it for me please?
• May 14th 2008, 06:42 PM
szpengchao
be careful
be careful. hav u seen that f'(a) < k < f'(b)

that is f', not f
• May 14th 2008, 07:30 PM
ThePerfectHacker
The intermediate value theorem for derivatives states that if $\displaystyle f:I\mapsto \mathbb{R}$ is a differenciable function then $\displaystyle f'$ has the intermediate value property on $\displaystyle I$, an open interval. Meaning if $\displaystyle a,b\in I$ with $\displaystyle a<b$ and $\displaystyle c$ lies (strictly) between $\displaystyle f'(a)$ and $\displaystyle f'(b)$ there exists $\displaystyle x_0\in (a,b)$ so that $\displaystyle f'(x_0) = c$.

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