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Darboux's theorem: every derivative has the intermediate-value property

Statement

If $I\subseteq\mathbb R$ is an interval and $f:I\to\mathbb R$ is differentiable, then $f'$ has the intermediate value property (The intermediate value property (Darboux property) of a function on an interval: the image of every subinterval is order-convex).

Facts & Assumptions

Given: $x<y$ in $I$ and a real $\lambda$ between $f'(x)$ and $f'(y)$ .

[L1]

Differentiability implies continuity; the closed bounded interval $[x,y]$ is compact; and a continuous real function on a nonempty compact set attains its extrema (A function differentiable at $c$ is continuous at $c$ , Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact, Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value).

[L2]

An interior extremum of a differentiable function has derivative $0$ (Fermat's interior extremum theorem: if $f$ has a local extremum at a point $c$ interior to its domain and is differentiable at $c$ , then $f'(c) = 0$ ).

[L3]

Derivatives obey the algebra rules, and the derivative of $t\mapsto t$ is $1$ (Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ , For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

Proof

technique · cases

1.1

If $\lambda=f'(x)$ or $\lambda=f'(y)$ , choose that endpoint.

assume-case endpointgiven

1.2

Suppose $f'(x)<\lambda<f'(y)$ , and define $h(t)=f(t)-\lambda t$ on $[x,y]$ . Then $h'(x)<0<h'(y)$ .

assume-case increasingL3algebra

1.3

If instead $f'(y)<\lambda<f'(x)$ , apply the preceding argument to $-h$ , obtaining an interior extremum of $h$ .

assume-case decreasingL1L3

2.1

For sufficiently small positive $s,t$ , the derivative inequalities give $h(x+s)<h(x)$ and $h(y-t)<h(y)$ . Hence a minimum of $h$ on $[x,y]$ occurs at an interior point $c$ .

step 1.2L1choose

3.1

In either strict-order case, Fermat gives $h'(c)=0$ , hence $f'(c)=\lambda$ . Together with the endpoint case, every intermediate value is attained.

step 2.1step 1.3L2L3cases-exhaustive∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 93 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, Taylor's theorem and related calculus (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 lecture notes (standard reference, not scraped)
J. Lebl, Basic Analysis I, Mean value theorem (standard reference, not scraped)
Colgate University MATH 323, Chapter 5 notes (standard reference, not scraped)