# Rolle's theorem

## 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

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Math_QED
Homework Helper
2019 Award
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.

Last edited:
• mathwonk
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

Math_QED
Homework Helper
2019 Award
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