Derivative of differentiable function satisfies intermediate value property

Statement

Two-sided version

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 $f^{\prime }(c)=w$.

Endpoint one-sided version

Suppose $f$ is a function defined on an interval $[a,b]$. Suppose:

• $f^{\prime }$ (the derivative of $f$) exists on the open interval $(a,b)$.
• The right hand derivative $f{\text{'}}_{+}(a)$ exists
• The left hand derivative $f{\text{'}}_{-}(b)$ exists

Suppose further that $f{\text{'}}_{+}(a)\ne f{\text{'}}_{-}(b)$ and $w$ is a real number strictly between these two numbers. Then, 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. Differentiable implies continuous
2. Intermediate value theorem
3. 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 (3).

Proof details for one-sided endpoint version

The two-sided version follows from the one-sided endpoint version, so we only prove the latter.

Given: $f$ is a function defined on an interval $[a,b]$. Suppose:

• $f^{\prime }$ (the derivative of $f$) exists on the open interval $(a,b)$.
• The right hand derivative $f{\text{'}}_{+}(a)$ exists
• The left hand derivative $f{\text{'}}_{-}(b)$ exists

Further, $f{\text{'}}_{+}(a)\ne f{\text{'}}_{-}(b)$ and $w$ is a real number strictly between these two numbers.

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. Note that we use $f{\text{'}}_{+}$ for the right hand derivative and $f{\text{'}}_{-}$ for the left hand derivative:

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

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 $(1/2,1)$.	Fact (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 $f{\text{'}}_{+}(a)$, which is $g(0)$.
3	$g$ is continuous at $t=1/2$	Fact (1)			We take the limit by plugging $t=1/2$ in the left side and right side definition and check. We again use the continuity of $f$ by Fact (1).
4	$g$ is left continuous at $t=1$	definition of derivative as a limit of a difference quotient			[SHOW MORE] 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 by definition is the left hand derivative $f{\text{'}}_{-}(b)$, 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 (2)	$w$ 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 (3)		Step (6)	Step-fact combination direct.