(Theorem) Intermediate value

For continuous functions

Theorem: Intermediate values for continuous functions

Let $f:[a,b]\to \mathbb{R}$ be continuous. Suppose $f(a)\le f(b)$, and $y_0\in \mathbb{R}$ has $f(a)\le y_0\le f(b)$. Then there exists some $c\in [a,b]$ so that $f(x)=y_0$.

As a corollary, we see that The function $f:[0,\infty )\to [0,\infty )$ defined as $f(x)=x^n$ is a bijection, so its inverse $g(x)=x^{1/n}$ is defined.

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For derivatives

A function that has a derivative defined on its entire domain is not necessarily continuous on that domain. However, a function that is a derivative shares an important property with a function that is continuous: we can assume intermediate values.

Theorem: Intermediate values for derivatives

Let $f:[a,b]\to \mathbb{R}$ be differentiable. If $f’(a)\le c\le f’(b)$ for some $c\in \mathbb{R}$, then there exists $x_0\in (a,b)$ such that $f’(x_0)=c$.

Proof from Modern Analysis I. Uses:

• (Proof pattern) Defining a new function

Define a new function $g:[a,b]\to \mathbb{R}$ by $g(x)=f(x)-cx$, so that $g’(x)=0$ if and only if $f’(x)=c$. Since $g$ is differentiable, $g$ is continuous, and it follows from the extreme value theorem that $g$ attains a minimum at some $x\in [a,b]$. Further, at the boundary, we have

$$g^{\prime }(a)=f^{\prime }(a)-c<0g^{\prime }(b)=f^{\prime }(b)-c>0,$$

so the minimum cannot be attained on either $a,b$. Then $x\in (a,b)$ with $f’(x)=c$, as desired. $\square$

wip Why minimum?

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Notes

• Special case of no-retract theorem from algebraic topology.