Mean Value Theorem

{{calculus}}
Let ''f'' : [''a'', ''b''] → '''R''' be a [[continuous function]] on the closed [[interval (mathematics)|interval]] [''a'', ''b''], and [[derivative|differentiable]] on the open interval (''a'', ''b''), where {{nowrap|''a'' < ''b''.}} Then there exists some ''c'' in (''a'', ''b'') such that
::$f\; \text{'}\; (c)\; =\; \backslash frac\{f(b)\; -\; f(a)\}\{b\; -\; a\}.$

The mean value theorem is a generalization of [[Rolle's theorem]], which assumes ''f''(''a'') = ''f''(''b''), so that the right-hand side above is zero.

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that ''f'' : [''a'', ''b''] → '''R''' is [[continuous function|continuous]] on [''a'', ''b''], and that for every ''x'' in (''a'', ''b'') the [[limit of a function|limit]]

:$\backslash lim\_\{h\backslash to\; 0\}\backslash frac\{f(x+h)-f(x)\}\{h\}$

exists as a finite number or equals +∞ or −∞. If finite, that limit equals ''f' ''(''x''). An example where this version of the theorem applies is given by the real-valued [[cube root]] function mapping ''x'' to ''x''^1/3, whose [[derivative]] tends to infinity at the origin.

Note that the theorem is false if a differentiable function is complex-valued instead of real-valued. Indeed, define $f(x)\; =\; e^\{ix\}$ for all real ''x''. Then
:$f(2\backslash pi)\; -\; f(0)\; =\; 0\; =\; 0\; (2\backslash pi\; -\; 0)$,
while
:$|f\; \text{'}(x)|\; =\; 1$.