# Intermediate value theorem

## Homework Statement

Let ##a,b\in \mathbb{R}##, ##a<b## and let ##f## be a continuous real valued function on ##[a,b]##. Prove that if ##f## is one-one then ##f([a,b])## is either ##[f(a),f(b)]## or ##[f(b),f(a)].##

2. The attempt at a solution

Suppose ##f(a) < f(b)## then by IVT we have if ##x\in(a,b)## then ##f(x) \in [f(a),f(b)]##. Arguing by contradiction suppose ##f(x) < f(a)## or ##f(x)>f(b)##. If ##f(x) <f(a)## then by IVT ##f([x,b])## contains all points between ##f(x)## and ##f(b)## and contains ##f(a)## contradiction that ##f## is one to one.

Why does that contradict the assumption that ##f## is one to one?

Dick
Homework Helper

## Homework Statement

Let ##a,b\in \mathbb{R}##, ##a<b## and let ##f## be a continuous real valued function on ##[a,b]##. Prove that if ##f## is one-one then ##f([a,b])## is either ##[f(a),f(b)]## or ##[f(b),f(a)].##

2. The attempt at a solution

Suppose ##f(a) < f(b)## then by IVT we have if ##x\in(a,b)## then ##f(x) \in [f(a),f(b)]##. Arguing by contradiction suppose ##f(x) < f(a)## or ##f(x)>f(b)##. If ##f(x) <f(a)## then by IVT ##f([x,b])## contains all points between ##f(x)## and ##f(b)## and contains ##f(a)## contradiction that ##f## is one to one.

Why does that contradict the assumption that ##f## is one to one?

Draw a picture. Suppose f(a)<=f(b) and x is in (a,b) but f(x)<f(a). Doesn't that mean that there is a y in [x,b] such that f(y)=f(a)?

• 1 person
Thank you! I see my mistake in my reasoning now.

Last edited:
Dick
Homework Helper
Thank you! I see my mistake in reasoning now.

I don't think there was any real mistake. Are you clear how on how f(y)=f(a) would contradict one-to-one?

Yup, because we already know ##a<x<b## and if ##f(a)\in f([x,b])## we have that ##x\le y\le b## thus ##f(y) = f(a)## which contradicts one to one since ##y\ne a##.

Last edited:
Dick
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