A question regarding continuous function on a closed interval

[aalma]

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.

 

[pasmith]

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 $$g: [a,b] \to \mathbb{R} : x \mapsto (f(x) - f(a))(f(b) - f(x))$$ the obvious candidate?

 

[aalma]

Isn't $$g: [a,b] \to \mathbb{R} : x \mapsto (f(x) - f(a))(f(b) - f(x))$$ 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 $g(c)<0$?

 

[fresh_42]

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 $g(x).$

 

[aalma]

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?)

 

[fresh_42]

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.

 

[Steve4Physics]

$(f(c) - f(a))((f)(b) - f(c)) <0$ 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.

 

[pasmith]

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

 

[julian]

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.

 

[mathwonk]

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.