Let ${\displaystyle I}$  be a closed interval, ${\displaystyle f\colon I\to \mathbb {R} }$  be a real-valued differentiable function. Then ${\displaystyle f'}$  has the intermediate value property: If ${\displaystyle a}$  and ${\displaystyle b}$  are points in ${\displaystyle I}$  with ${\displaystyle a<b}$ , then for every ${\displaystyle y}$  between ${\displaystyle f'(a)}$  and ${\displaystyle f'(b)}$ , there exists an ${\displaystyle x}$  in ${\displaystyle [a,b]}$  such that ${\displaystyle f'(x)=y}$ .^[1][2][3]

Proof 1. The first proof is based on the extreme value theorem.

If ${\displaystyle y}$  equals ${\displaystyle f'(a)}$  or ${\displaystyle f'(b)}$ , then setting ${\displaystyle x}$  equal to ${\displaystyle a}$  or ${\displaystyle b}$ , respectively, gives the desired result. Now assume that ${\displaystyle y}$  is strictly between ${\displaystyle f'(a)}$  and ${\displaystyle f'(b)}$ , and in particular that ${\displaystyle f'(a)>y>f'(b)}$ . Let ${\displaystyle \varphi \colon I\to \mathbb {R} }$  such that ${\displaystyle \varphi (t)=f(t)-yt}$ . If it is the case that ${\displaystyle f'(a)<y<f'(b)}$  we adjust our below proof, instead asserting that ${\displaystyle \varphi }$  has its minimum on ${\displaystyle [a,b]}$ .

Since ${\displaystyle \varphi }$  is continuous on the closed interval ${\displaystyle [a,b]}$ , the maximum value of ${\displaystyle \varphi }$  on ${\displaystyle [a,b]}$  is attained at some point in ${\displaystyle [a,b]}$ , according to the extreme value theorem.

Because ${\displaystyle \varphi '(a)=f'(a)-y>0}$ , we know ${\displaystyle \varphi }$  cannot attain its maximum value at ${\displaystyle a}$ . (If it did, then ${\displaystyle (\varphi (t)-\varphi (a))/(t-a)\leq 0}$  for all ${\displaystyle t\in (a,b]}$ , which implies ${\displaystyle \varphi '(a)\leq 0}$ .)

Likewise, because ${\displaystyle \varphi '(b)=f'(b)-y<0}$ , we know ${\displaystyle \varphi }$  cannot attain its maximum value at ${\displaystyle b}$ .

Therefore, ${\displaystyle \varphi }$  must attain its maximum value at some point ${\displaystyle x\in (a,b)}$ . Hence, by Fermat's theorem, ${\displaystyle \varphi '(x)=0}$ , i.e. ${\displaystyle f'(x)=y}$ .

Proof 2. The second proof is based on combining the mean value theorem and the intermediate value theorem.^[1][2]

Define ${\displaystyle c={\frac {1}{2}}(a+b)}$ . For ${\displaystyle a\leq t\leq c,}$  define ${\displaystyle \alpha (t)=a}$  and ${\displaystyle \beta (t)=2t-a}$ . And for ${\displaystyle c\leq t\leq b,}$  define ${\displaystyle \alpha (t)=2t-b}$  and ${\displaystyle \beta (t)=b}$ .

Thus, for ${\displaystyle t\in (a,b)}$  we have ${\displaystyle a\leq \alpha (t)<\beta (t)\leq b}$ . Now, define ${\displaystyle g(t)={\frac {(f\circ \beta )(t)-(f\circ \alpha )(t)}{\beta (t)-\alpha (t)}}}$  with ${\displaystyle a<t<b}$ . ${\displaystyle \,g}$  is continuous in ${\displaystyle (a,b)}$ .

Furthermore, ${\displaystyle g(t)\rightarrow {f}'(a)}$  when ${\displaystyle t\rightarrow a}$  and ${\displaystyle g(t)\rightarrow {f}'(b)}$  when ${\displaystyle t\rightarrow b}$ ; therefore, from the Intermediate Value Theorem, if ${\displaystyle y\in ({f}'(a),{f}'(b))}$  then, there exists ${\displaystyle t_{0}\in (a,b)}$  such that ${\displaystyle g(t_{0})=y}$ . Let's fix ${\displaystyle t_{0}}$ .

From the Mean Value Theorem, there exists a point ${\displaystyle x\in (\alpha (t_{0}),\beta (t_{0}))}$  such that ${\displaystyle {f}'(x)=g(t_{0})}$ . Hence, ${\displaystyle {f}'(x)=y}$ .

A Darboux function is a real-valued function ƒ which has the "intermediate value property": for any two values a and b in the domain of ƒ, and any y between ƒ(a) and ƒ(b), there is some c between a and b with ƒ(c) = y.^[4] By the intermediate value theorem, every continuous function on a real interval is a Darboux function. Darboux's contribution was to show that there are discontinuous Darboux functions.

Every discontinuity of a Darboux function is essential, that is, at any point of discontinuity, at least one of the left hand and right hand limits does not exist.

An example of a Darboux function that is discontinuous at one point is the topologist's sine curve function:

${\displaystyle x\mapsto {\begin{cases}\sin(1/x)&{\text{for }}x\neq 0,\\0&{\text{for }}x=0.\end{cases}}}$ 

By Darboux's theorem, the derivative of any differentiable function is a Darboux function. In particular, the derivative of the function ${\displaystyle x\mapsto x^{2}\sin(1/x)}$  is a Darboux function even though it is not continuous at one point.

An example of a Darboux function that is nowhere continuous is the Conway base 13 function.

Darboux functions are a quite general class of functions. It turns out that any real-valued function ƒ on the real line can be written as the sum of two Darboux functions.^[5] This implies in particular that the class of Darboux functions is not closed under addition.

A strongly Darboux function is one for which the image of every (non-empty) open interval is the whole real line. The Conway base 13 function is again an example.^[4]

• This article incorporates material from Darboux's theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
• "Darboux theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]