A question regarding continuous function on a closed interval

##(f(c) - f(a))((f)(b) - f(c)) <0## tells us that there are two cases:

##f(c) >f(a), f(b) ##
##f(c) <f(a), f(b) ##.
I guess we need to define a new function here that let us use the Rolle's theorem..
But it is not clear enough how to do so.

aalma said:

Homework Statement: Let ##f## be continuous on ##[a, b] ## and differentiable on ##(a, b)## and assume there is ##c\in(a, b) ## such that ##(f(c) - f(a))(f(b) - f(c)) <0## then there exists ##t\in(a, b) ## such that ##f'(t) =0##.
Relevant Equations: ##(f(c) - f(a))((f)(b) - f(c)) <0##

##(f(c) - f(a))((f)(b) - f(c)) <0## tells us that there are two cases:

##f(c) >f(a), f(b) ##
##f(c) <f(a), f(b) ##.
I guess we need to define a new function here that let us use the Rolle's theorem..
But it is not clear enough how to do so.

Isn't [tex]
g: [a,b] \to \mathbb{R} : x \mapsto (f(x) - f(a))(f(b) - f(x))[/tex] the obvious candidate?

pasmith said:

Isn't [tex]
g: [a,b] \to \mathbb{R} : x \mapsto (f(x) - f(a))(f(b) - f(x))[/tex] the obvious candidate?

Yeah, nut then ##g(a)=0=g(b)## and since g is continuos on [a,b] and differentiable on (a,b) so by Roll's theorem there is ##t \in (a,b): g'(t)=0## but then how to use that ##g(c)<0##?

aalma said:

Yeah, nut then ##g(a)=0=g(b)## and since g is continuos on [a,b] and differentiable on (a,b) so by Roll's theorem there is ##t \in (a,b): g'(t)=0## but then how to use that ##g(c)<0##?

You need ##f'(t)=0.## Differentiate ##g(x).##

fresh_42 said:

You need ##f'(t)=0.## Differentiate ##g(x).##

Right, this gives ##g'(x)=f'(x)(f(a)-2f(x)+f(b))##. ##g'(t)=0## but how this says that ##f'(t)## must be 0? (How we use the assumption abou c?)

I haven't done the problem.

The idea seems to be to prove that there is at least a minimum or a maximum between ##a## and ##b.## That's Weierstraß's extreme value theorem. It is proven with sequences. I assume we are supposed to prove it differently. The existence of ##c## guarantees at least that ##f## isn't a constant function. We can further assume that ##f(c) > f(a)## and ##f(c) > f(b)## as the other case is very likely along the same line of proof.

Rolle gives us - excuse me that I check your calculation -
\begin{align*}
g'(x)&=f'(x)(f(b)-f(x)) -f'(x)(f(x)-f(a))=f'(x)(f(b)-2f(x)+f(a))\\
g'(\xi_1)&=0=f'(\xi_1)(f(b)-2f(\xi)+f(a))
\end{align*}
If ##f'(\xi_1)=0## then we are done. Otherwise,
\begin{align*}
0&=f(b)-2f(\xi_1)+f(a) \Longrightarrow f(\xi_1)=\dfrac{f(b)+f(a)}{2} <f(c)
\end{align*}
I would now repeat this process on ##[a,\xi_1].## We either find a value ##f'(\xi_n)=0## or ##\displaystyle{\lim_{n \to \infty}\xi_n=a.} ## This means ##\displaystyle{\lim_{n \to \infty}f(\xi_n)=f(a)} ## by continuity of ##f##.

We must now show that this cannot happen. Something must go wrong when ##c\not\in [a,\xi_n]## in our process anymore.

aalma said:

##(f(c) - f(a))((f)(b) - f(c)) <0## tells us that there are two cases:

##f(c) >f(a), f(b) ##
##f(c) <f(a), f(b) ##.

Here’s a (non-mathematician’s) suggestion...

Consider, say, the first case where ##f(c) >f(a), f(b)##. (This corresponds to the existance of one or more maxima in the interval.)

For the purpose of explanation, suppose that ##f(a)>f(b)##.

Since ##f(x)## is continuous, there is some value of ##x## (say ##x=d##) such that ##f(d) = f(a)##.
,
##c## lies in the interval ##(a, d)##.

For the interval ##(a, d)##, ##f(x)## has equal endpoints, so Rolle’s theorem applies directly.

aalma said:

Right, this gives ##g'(x)=f'(x)(f(a)-2f(x)+f(b))##. ##g'(t)=0## but how this says that ##f'(t)## must be 0? (How we use the assumption abou c?)

If [itex]g'(\xi) = 0[/itex] then either [itex]f'(\xi) = 0[/itex] or [itex]f(\xi) = (f(b) + f(a))/2[/itex]. in the first case we are done. What is the value of [itex]g(\xi)[/itex] in the second case? Does the fact that [itex]g(c) < 0[/itex] allow you to find a smaller interval on which to apply Rolle's Thorem again?

Steve4Physics said:

Here’s a (non-mathematician’s) suggestion...

Consider, say, the first case where ##f(c) >f(a), f(b)##. (This corresponds to the existance of one or more maxima in the interval.)

For the purpose of explanation, suppose that ##f(a)>f(b)##.

Since ##f(x)## is continuous, there is some value of ##x## (say ##x=d##) such that ##f(d) = f(a)##.
,
##c## lies in the interval ##(a, d)##.

For the interval ##(a, d)##, ##f(x)## has equal endpoints, so Rolle’s theorem applies directly.

I think you would apply the Intermediate Value Theorem in this context. It states:

If ##f## is a continuous function whose domain contains the interval ##[c, b]##, then it takes on any given value between ##f(c)## and ##f(b)## at some point within the interval.

I agree to use IVT. This is one of my favorite applications of Rolle. I.e. to show that a differentiable function that changes direction must have zero derivative somewhere in between. It is a bit tedious to treat all cases, but the point is that a continuous function that changes direction on an interval must take the same value at some two distinct points.

Note the hypothesis here implies the function is not monotone on (a,b).

A nice corollary is that a differentiable function on an interval, with derivative never zero, must be monotone. People usually use MVT for this but this shows Rolle is enough.