# Difference between revisions of "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 $\Delta f(a,b)$. More explicitly:

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

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).

## 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]
4 $g$ is differentiable on $(a,b)$ Fact (2) $f$ is differentiable on $(a,b)$ Steps (1), (2) [SHOW MORE]
5 $g(a) = g(b) = 0$ Steps (1), (2) [SHOW MORE]
6 There exists $c \in (a,b)$ such that $g'(c) = 0$. Fact (3) Steps (3), (4), (5) [SHOW MORE]
7 For the $c$ obtained in step (6), $f'(c) = h'(c)$ Fact (4) Steps (2), (6) From Step (2), $g = f - h$. Differentiating both sides by Fact (4), we get $g' = f' - h'$ on $(a,b)$. Since $g'(c) = 0$, we obtain that $f'(c) = h'(c)$.</toggledisplay>
8 $h'(x) = \frac{f(b) - f(a)}{b - a}$ for all $x$. In particular, $h'(c) = \frac{f(b) - f(a)}{b - a}$. Step (1) Differentiate the expression for $h(x)$ from Step (1).
9 $f'(c) = \frac{f(b) - f(a)}{b - a}$ Steps (7), (8) Step-combination direct