Rolle's Theorem: Proving Fixed & Non-Fixed Cases

[sergey_le]

Homework Statement

Let function ƒ be Continuous in the compact interval [a,b].
if f(a)=f(b), then f is not Injective in the open interval (a,b)

Relevant Equations

Rolle's theorem

It all makes sense to me, but I don't know how to formalize it nicely.
I wanted to divide it into two cases.
First case where f is fixed in the segment.
And a second case where f is not fixed in the segment.
But I don't know how to prove it for the case where f i is not fixed

 

[member 587159]

Rolle's theorem won't be useful here because you are not given that $f$ is differentiable. This looks like another application of the intermediate value theorem.

Hint: Assume to the contrary that $f$ is injective on $]a,b[$. Look at the point $x= (a+b)/2$ (or any other point in $]a,b[$, but the middle of the interval seems like the canonical choice). You know that $f(x) \neq f(a)$, so either $f(x) > f(a)$ or $f(x) < f(a)$. Assume without loss of generality $f(x) > f(a)$.

The intuition now is that to get from $f(a)$ to $f(x)$ you cross the interval $[f(a), f(x)]$ and to get from $f(x)$ to $f(b)$ you cross the interval $[f(b), f(x)]$ again with other $x$-values. The intermediate value theorem tells that this is impossible because we assumed that $f$ is injective.

Try to make this argument formal now.

 

[sergey_le]

Rolle's theorem won't be useful here because you are not given that $f$ is differentiable. This looks like another application of the intermediate value theorem.

Hint: Assume to the contrary that $f$ is injective on $]a,b[$. Look at the point $x= (a+b)/2$ (or any other point in $]a,b[$, but the middle of the interval seems like the canonical choice). You know that $f(x) \neq f(a)$, so either $f(x) > f(a)$ or $f(x) < f(a)$. Assume without loss of generality $f(x) > f(a)$.

The intuition now is that to get from $f(a)$ to $f(x)$ you cross the interval $[f(a), f(x)]$ and to get from $f(x)$ to $f(b)$ you cross the interval $[f(b), f(x)]$ again with other $x$-values. The intermediate value theorem tells that this is impossible because we assumed that $f$ is injective.

Try to make this argument formal now.

Ok, I sort of understand what you want me to do.
You want me to divide the interval [a,b] Into two parts With the help of a midpoint c.
And say that For any t that is between f(a) and f(c) Exists x1∈[a,c] so that f(x1)=t , Then switch the point f(a) whit f(b) And say that ∃x2∈[c,b] so that f(x2)=t .
But there can be a situation where x1=c=x2 and than f(x1) and f(x2) That's the same point.
I hope you understand me

 

[member 587159]

Ok, I sort of understand what you want me to do.
You want me to divide the interval [a,b] Into two parts With the help of a midpoint c.
And say that For any t that is between f(a) and f(c) Exists x1∈[a,c] so that f(x1)=t , Then switch the point f(a) whit f(b) And say that ∃x2∈[c,b] so that f(x2)=t .
But there can be a situation where x1=c=x2 and than f(x1) and f(x2) That's the same point.
I hope you understand me

Restrict your $x_1$ choices. Choose them in $]a,c[$ and not in $[a,c]$. Similarly for the other interval.

 

[sergey_le]

Restrict your $x_1$ choices. Choose them in $]a,c[$ and not in $[a,c]$. Similarly for the other interval.

Thanks