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Mean value theorem
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==Implications== '''Theorem 1:''' Assume that <math>f</math> is a continuous, real-valued function, defined on an arbitrary interval <math>I</math> of the real line. If the derivative of <math>f</math> at every [[interior (topology)|interior point]] of the interval <math>I</math> exists and is zero, then <math>f</math> is [[constant function|constant]] on <math>I</math>. '''Proof:''' Assume the derivative of <math>f</math> at every [[interior (topology)|interior point]] of the interval <math>I</math> exists and is zero. Let <math>(a, b)</math> be an arbitrary open interval in <math>I</math>. By the mean value theorem, there exists a point <math>c</math> in <math>(a, b)</math> such that :<math>0=f'(c)=\frac{f(b)-f(a)}{b-a}.</math> This implies that {{nowrap|1=<math>f(a) = f(b)</math>}}. Thus, <math>f</math> is constant on the interior of <math>I</math> and thus is constant on <math>I</math> by continuity. (See below for a multivariable version of this result.) '''Remarks:''' * Only continuity of <math>f</math>, not differentiability, is needed at the endpoints of the interval <math>I</math>. No hypothesis of continuity needs to be stated if <math>I</math> is an [[open interval]], since the existence of a derivative at a point implies the continuity at this point. (See the section [[Derivative#Continuity and differentiability|continuity and differentiability]] of the article [[derivative]].) * The differentiability of <math>f</math> can be relaxed to [[one-sided derivatives|one-sided differentiability]], a proof is given in the article on [[semi-differentiability]]. '''Theorem 2:''' If <math>f'(x) = g'(x)</math> for all <math>x</math> in an interval <math>(a, b)</math> of the domain of these functions, then <math>f - g</math> is constant, i.e. <math>f = g + c</math> where <math>c</math> is a constant on <math>(a, b)</math>. '''Proof:''' Let <math>F(x) = f(x) - g(x)</math>, then <math>F'(x)=f'(x)-g'(x)=0</math> on the interval <math>(a, b)</math>, so the above theorem 1 tells that <math>F(x) = f(x) - g(x)</math> is a constant <math>c</math> or <math>f = g + c</math>. '''Theorem 3:''' If <math>F</math> is an antiderivative of <math>f</math> on an interval <math>I</math>, then the most general antiderivative of <math>f</math> on <math>I</math> is <math>F(x) + c</math> where <math>c</math> is a constant. '''Proof:''' It directly follows from the theorem 2 above.
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