Mean value theorem ($f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$)

The mean value theorem says that an average rate of change is attained as an instantaneous rate somewhere between the endpoints. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Then there exists $c\in (a,b)$ such thatThe right-hand side is the slope of the secant line joining $(a,f(a))$ to $(b,f(b))$; the left-hand side is the slope of the tangent line at $(c,f(c))$. The derivative need not be continuous, and the function need not be differentiable at the endpoints. ^[ref]

For the polynomial $f(x)=x^2$ on $[1,3]$, the secant line has slope $(9-1)/(3-1)=4$. Since the derivative is $f^{\prime }(x)=2x$, the required interior point is $c=2$. The tangent line there is $y=4+4(x-2)=4x-4$, whereas the secant line is $y=1+4(x-1)=4x-3$. Their equal slopes make them parallel, but they are distinct straight lines: the theorem equates rates, not heights.

The practical strength is that the unknown point $c$ can disappear from an estimate. Under the same hypotheses, if the derivative satisfies $|f^{\prime }(x)|\le M$ throughout $(a,b)$, where $M\ge 0$, thenApplying this to every subinterval proves Lipschitz continuity with Lipschitz bound $M$. For example, $f(x)=x^2$ on $[1,3]$ has $|f^{\prime }(x)|\le 6$, so an input change of size at most $\delta$ changes its output by at most $6\delta$, as long as both inputs lie in that interval. A local derivative bound therefore controls finite changes without computing an integral. The sign of the derivative similarly gives the derivative criterion for monotonicity.

Taking $M=0$ shows that a function with zero derivative throughout a real interval is a constant function, provided it is continuous at any included endpoints. The interval condition is essential: on the domain of a function $(-2,-1)\cup (1,2)$, the function equal to zero on the left interval and one on the right interval has zero derivative everywhere. It is a locally constant function, yet not a constant function on the whole domain of a function. The mean value theorem cannot cross the gap.

A concise proof subtracts the secant line: set $m=(f(b)-f(a))/(b-a)$ and $h(x)=f(x)-f(a)-m(x-a)$. Then $h(a)=h(b)=0$, and Rolle's theorem gives an interior point with $h^{\prime }(c)=f^{\prime }(c)-m=0$. The geometric idea is to remove the average trend and find a horizontal tangent line in what remains.

A differentiable function on an interval whose derivative is identically zero is a constant function. Indeed, for any two arguments $x<y$, the mean value theorem on $[x,y]$ gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. For an interval with endpoints included, continuity at those endpoints is required. A disconnected domain may have different constants on different components.

For $a<b$ and real-valued functions $f,g$ that are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, there exists $\xi \in (a,b)$ withApply Rolle's theorem to the linear combination $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, which has equal endpoint values. The displayed product identity permits zero derivatives. For the ratio form, additionally assume $g^{\prime }(x)\ne 0$ for every $x\in (a,b)$. Then Rolle's theorem ensures $g(b)\ne g(a)$, so both denominators below are nonzero:Neither derivative needs to be continuous, and $f^{\prime }$ may vanish. Choosing $g(x)=x$ recovers the mean value theorem; the ratio form also gives the zero-over-zero endpoint case of L'Hôpital's rule. ^[ref]

If $f$ is continuous at an interval endpoint, differentiable in its interior, and has a finite one-sided derivative there, then its secant slope and tangent slope are arbitrarily close at some interior point. Apply the mean value theorem on a short endpoint interval; both the secant at its far endpoint and the secant at the resulting intermediate point approach the same one-sided derivative. No continuity of the derivative is required.