Lagrange mean value theorem

Statement

Suppose $f$ is a function defined on a closed interval $[a,b]$ (with $a<b$) such that the following two conditions hold:

1. $f$ is a continuous function on the closed interval $[a,b]$ (i.e., it is right continuous at $a$, left continuous at $b$, and two-sided continuous at all points in the open interval $(a,b)$).
2. $f$ is a differentiable function on the open interval $(a,b)$, i.e., the derivative exists at all points in $(a,b)$. Note that we do not require the derivative of $f$ to be a continuous function.

Then, there exists $c$ in the open interval $(a,b)$ such that the derivative of $f$ at $c$ equals the difference quotient $\mathrm{\Delta }f(a,b)$. More explicitly:

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

Geometrically, this is equivalent to stating that the tangent line to the graph of $f$ at $c$ is parallel to the chord joining the points $(a,f(a))$ and $(b,f(b))$.

Note that the theorem simply guarantees the existence of $c$, and does not give a formula for finding such a $c$ (which may or may not be unique).

Related facts

• Rolle's theorem
• Zero derivative implies locally constant
• Fundamental theorem of calculus
• Positive derivative implies increasing
• Increasing and differentiable implies nonnegative derivative
• Derivative of differentiable function on interval satisfies intermediate value property

Facts used

1. Continuous functions form a vector space
2. Differentiable functions form a vector space
3. Rolle's theorem
4. Differentiation is linear

Proof

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the function $h(x):=\frac{f(b)-f(a)}{b-a}\cdot x+\frac{bf(a)-af(b)}{b-a}$. Then, $h$ is a linear (and hence a continuous and differentiable) function with $h(a)=f(a)$ and $h(b)=f(b)$				Just plug in and check. Secretly, we obtained $h$ by trying to write the equation of the line joining the points $(a,f(a))$ and $(b,f(b))$.
2	Define $g=f-h$ on $[a,b]$, i.e., $g(x):=f(x)-h(x)$.
3	$g$ is continuous on $[a,b]$	Fact (1)	$f$ is continuous on $[a,b]$	Steps (1), (2)	[SHOW MORE] By Step (2), $g=f-h$. By Step (1), $h$ is linear and hence continuous on $[a,b]$. Thus, by Fact (1), $g=f-h$, being the difference of two continuous functions, is continuous on $[a,b]$.
4	$g$ is differentiable on $(a,b)$	Fact (2)	$f$ is differentiable on $(a,b)$	Steps (1), (2)	[SHOW MORE] By Step (2), $g=f-h$. By Step (1), $h$ is linear and hence differentiable on $(a,b)$. Thus, by Fact (1), $g=f-h$, being the difference of two differentiable functions, is differentiable on $(a,b)$.
5	$g(a)=g(b)=0$			Steps (1), (2)	[SHOW MORE] $h(a)=f(a)$ yields $g(a)=0$. $h(b)=f(b)$ yields $g(b)=0$.
6	There exists $c\in (a,b)$ such that $g^{\prime }(c)=0$.	Fact (3)		Steps (3), (4), (5)	[SHOW MORE] Steps (3), (4), (5) together show that $g$ satisfies the conditions of Rolle's theorem on the interval $[a,b]$, hence we get the conclusion of the theorem.
7	For the $c$ obtained in step (6), $f^{\prime }(c)=h^{\prime }(c)$	Fact (4)		Steps (2), (6)	From Step (2), $g=f-h$. Differentiating both sides by Fact (4), we get $g^{\prime }=f^{\prime }-h^{\prime }$ on $(a,b)$. Since $g^{\prime }(c)=0$, we obtain that $f^{\prime }(c)=h^{\prime }(c)$.</toggledisplay>
8	$h^{\prime }(x)=\frac{f(b)-f(a)}{b-a}$ for all $x$. In particular, $h^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$.			Step (1)	Differentiate the expression for $h(x)$ from Step (1).
9	$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$			Steps (7), (8)	Step-combination direct