Intermediate Value Theorem for derivatives

Theorem 4.2.13 (Intermediate Value Theorem for derivative)

Let $f:I\to\mathbb{R}$ be differentiable function on the interval $I=[a,b]$. Then given $a,b\in I$ with $a<b$ and $f^\prime(a) < \lambda <f^\prime(b)$, there exists $c\in(a,b)$ such that $f^\prime(c) = \lambda$.

The following proof is from Lars Oslen [1] <proof> Define continuous functions $f_a, f_b: [a,b]\to\mathbb{R}$ by

\(\begin{align} f_a(t)&= \begin{cases} f^\prime(a)\quad &( t = a)\\ \frac{f(a) -f(t)}{a-t}\quad &( t\neq a) \end{cases} \\ f_b(t) &= \begin{cases} f^\prime(b) \quad&(t=b)\\ \frac{f(t)-f(b)}{t-b}\quad&(t\neq b) \end{cases} \end{align}\) It follows that $f_a(a) = f^\prime(a), f_a(b)=f_b(a)$, and $f_b(b) = f^\prime(b)$. Hence $\lambda$ lies between $f_a(a)$ and $f_a(b)$ or $\lambda$ lies between $f_b(a)$ and $f_b(b)$.

If $\lambda$ lies between $f_a(a)$ and $f_a(b)$, then by continuity of $f_a$ there exists $s\in (a,b]$ with

\[\lambda = f_a(s) = \frac{f(s)-f(a)}{s-a}\]

The mean value theorem ensures that there exists $c\in [a,s]$ such that

\[\lambda = \frac{f(s)-f(a)}{s-a}=f^\prime(c)\]

If $\lambda$ lies between $f_b(a)$ and $f_b(b)$, then an analogous argument shows that there exists $s\in [a,b)$ such that

\(\lambda = \frac{f(b)-f(s)}{b-s} = f^\prime(c)\) where $c\in [s,b]$.

\[\tag*{$\square$}\]

Remark

Suppose that $f^\prime(c_1) = f^\prime(c_2) = 0$ with $c_1 < c_2$ and $f^\prime(x) \neq 0$ for all $x\in (c_1, c_2)$. It suffices to check the sign of the derivative at a single point in the interval $(c_1, c_2)$ to determine whether $f^\prime$ is positive or negative on the whole interval $(c_1,c_2)$.

$\because$ Suppose that $f^\prime(x_1)f^\prime(x_2) <0$ for some $x_1, x_2 \in (c_1,c_2)$. By Darboux theorem, there exists $c\in (x_1,x_2)$ such that $f(c)=0$, which contradicts to the assumption that $f^\prime(x)=0$ for all $x\in (c_1,c_2)$.

$\therefore f^\prime(x_1)f^\prime(x_2) >0$ for all $x_1, x_2 \in (c_1,c_2)$.

Theorem 5.2.14 (Inverse function theorem)

Let $f:I\to\mathbb{R}$ be differentiable on $I$ with $f^\prime (x) \neq 0$ for all $x\in I$. Then $f$ is one-to-one on $I$, the inverse function $f^{-1}$ is continuous, and differentiable on $I$ with

\[(f^{-1})^\prime(f(x)) = \frac{1}{f^\prime(x)}\]

for all $x\in I$.

<proof>

Since $f^\prime(x)\neq 0$ for all $x\in I$, $f^\prime$ is either positive or negative on $I$ by Darboux theorem. Without the loss of generality, assume that $f^\prime(x) >0$ for all $x\in I$. Then $f$ is strictly increasing on $I$, so $f$ is one-to-one function on $I$. Moreover $f^{-1}$ is continuous by theorem 4.4.12.

Since $f$ is continuous on $I$, $J=f(I)$ is an interval. Thus all $y\in J$ is a limit point of $J$. Now, let $y_0\in J=f(I)$ be given. Since $y_0$ is a limit point, we can define a sequence $(y_n)\in J$ with $y_n\to y_0$ but $y_n\neq y_0$.

Since $f^{-1}$ is continuous on $J$, $\displaystyle{\lim_{n\to\infty}f^{-1}(y_n)=f^{-1}(y_0)}$.

$\because$ \(\begin{align} \lim_{x\to p}f(x) = L \iff \lim_{n\to\infty}f(x_n) = L \text{ for all } \{x_n\} \text{ with } x_n \neq p \text{ s.t. }x_n\to p \text{ as } n\to\infty \end{align}\)

Then,

\[\lim_{n\to\infty}\frac{f^{-1}(y_n)-f^{-1}(y_0)}{y_n - y_0} = \lim_{n\to\infty}\frac{x_n - x_0}{f(x_n)-f(x_0)} = \frac{1}{f^\prime(x_0)}\]

$\therefore (f^{-1})^\prime(f(x)) = \frac{1}{f^\prime(x)}$, i.e., $f^{-1}(f(x_0)) = \frac{1}{f^\prime(x_0)}$.

$\therefore (f^{-1})^\prime(f(x))=\frac{1}{f^\prime(x)}$ for all $x\in I$. \(\tag*{$\square$}\)

Reference

[1] Oslen, Lars, A New Proof of darboux’s Theorem, The American Mathematical Society, Vol. 111, No. 8(Oct., 2004), pp. 713-715.

[2] Bhandari, Mukta Bahadur. “Another Proof of Darboux’s Theorem.” arXiv: History and Overview (2016): n. pag.

[3] Manfred Stoll, 『Introduction to Real Analysis』, Pearson