Derivative of differentiable function satisfies intermediate value property

Statement

Suppose $f$ is a differentiable function on an open interval $(p,q)$. Then, the derivative $f^{\prime }$ satisfies the intermediate value property on $(p,q)$: for $a<b$, both in $(p,q)$, and any value $w$ is strictly between $f^{\prime }(a)$ and $f^{\prime }(b)$, there exists $c\in (a,b)$ such that $f^{\prime }(c)=w$.

Related facts

• Intermediate value theorem: This states that any continuous function satisfies the intermediate value property.

Facts used

1. Lagrange mean value theorem

Proof

Proof idea

The proof idea is to find a difference quotient that takes the desired value intermediate between $f^{\prime }(a)$ and $f^{\prime }(b)$, then use Fact (1).

Proof details

Given: $f$ is a differentiable function on an open interval $(p,q)$. Suppose $a<b$, both in $(p,q)$, and suppose $w$ is strictly between $f^{\prime }(a)$ and $f^{\prime }(b)$,

To prove: There exists $c\in (a,b)$ such that $f^{\prime }(c)=w$.

Proof: Consider the function $g$ defined on the interval $[0,1]$ as follows. By $\mathrm{\Delta }f$ we denote the difference quotient:

$g(t):=\{\begin{array}{cc}f\text{'}(a),& t=0\\ (\mathrm{\Delta }f)(2tb+a(1-2t),a)=\frac{f(2tb+a(1-2t))-f(a)}{2t(b-a)},& 0<t<1/2\\ (\mathrm{\Delta }f)(b,a)=\frac{f(b)-f(a)}{b-a},& t=1/2\\ (\mathrm{\Delta }f)(b,(2-2t)a+b(2t-1))=\frac{f(b)-f((2-2t)a+b(2t-1))}{(2-2t)(b-a)},& 1/2<t<1\\ f\text{'}(b),& t=1\\ \end{array}$

Note that for $0<t<1$, $g(t)$ is a difference quotient between two points in $[a,b]$, at least one of which is one of the endpoints $a,b$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$g$ is continuous on $(0,1/2)$ and $(1/2,1)$.		$f$ is differentiable, hence continuous		Follows directly from continuity of $f$ and the nature of the expressions.
2	$g$ is right continuous at $t=0$	definition of derivative as a limit of a difference quotient		[SHOW MORE] The right hand limit at 0 is $\underset{t\to 0^{+}}{lim}(\mathrm{\Delta }f)(2tb+a(1-2t),a)$. Set $x=2tb+a(1-2t)$. The limit now becomes $\underset{x\to a^{+}}{lim}(\mathrm{\Delta }f)(x,a)=\underset{x\to a^{+}}{lim}\frac{f(x)-f(a)}{x-a}$ which is the definition of the right hand derivative. We are assuming the existence of a two-sided derivative $f^{\prime }(a)$, so the right hand derivative equals $f^{\prime }(a)$, which is $g(0)$.
3	$g$ is continuous at $t=1/2$				We plug $t=1/2$ in the left side and right side definition and check.
4	$g$ is left continuous at $t=1$	definition of derivative as a limit of a difference quotient			<toggledisplay>The left hand limit at 1 is $\underset{t\to 1^{-}}{lim}(\mathrm{\Delta }f)(b,(2-2t)a+b(2t-1))$. Put $x=(2-2t)a+b(2t-1)$ and plug in, and get $\underset{x\to b^{-}}{lim}(\mathrm{\Delta }f)(b,x)=\underset{x\to b^{-}}{lim}\frac{f(b)-f(x)}{b-x}=\underset{x\to b^{-}}{lim}\frac{f(x)-f(b)}{x-b}$, which is the definition of the left hand derivative. We are assuming the existence of a two-sided derivative $f^{\prime }(b)$, so the left hand derivative equals $f^{\prime }(b)$, which is $g(1)$.
5	$g$ is continuous on $[0,1]$.			Steps (1)-(4)	step-combination direct.
6	There exists $t\in (0,1)$ such that $g(t)=w$. In particular $w$ is a difference quotient between two points, both of them in $[a,b]$.	Fact (1)	$w$ is between $f^{\prime }(a)$ and $f^{\prime }(b)$.	Step (5)	By definition, $f^{\prime }(a)=g(0)$ and $f^{\prime }(b)=g(1)$. Step (5) and Fact (1) tell us that since $w$ is between these, there exists $t\in (0,1)$ such that $g(t)=w$.
7	There exists $c\in (a,b)$ such that $f^{\prime }(c)=w$.	Fact (2)		Step (6)	Step-fact combination direct.