Editing Mean value theorem (section)

== Cases where the theorem cannot be applied ==
All conditions for the mean value theorem are necessary:

# '''<math>\boldsymbol{f(x)} </math> is differentiable on <math>\boldsymbol{(a,b)} </math>''' 
# '''<math>\boldsymbol{f(x)} </math> is continuous on <math>\boldsymbol{[a,b]} </math>'''
# '''<math>\boldsymbol{f(x)} </math> is real-valued'''

When one of the above conditions is not satisfied, the mean value theorem is not valid in general, and so it cannot be applied.

[[File:Absolute Value.svg|thumb]]
The necessity of the first condition can be seen by the counterexample where the function <math>f(x)=|x| </math> on [-1,1] is not differentiable.

The necessity of the second condition can be seen by the counterexample where the function
<math display="block">f(x) = \begin{cases} 1, & \text{at }x=0 \\ 0, & \text{if }x\in( 0,1] \end{cases} </math>
satisfies criteria 1 since <math>f'(x)=0  </math> on <math>(0,1) </math> but not criteria 2 since 
<math display="block">\frac{f(1)-f(0)}{1-0}=-1 </math> and <math>-1\neq 0 =f'(x)  </math> for all <math>x\in (0,1) </math> so no such <math>c </math> exists.

The theorem is false if a differentiable function is complex-valued instead of real-valued. For example, if <math>f(x) = e^{xi}</math> for all real <math>x</math>, then
<math display="block">f(2\pi)-f(0)=0=0(2\pi-0)</math>
while <math>f'(x)\ne 0</math> for any real <math>x</math>.