# Changes

## Lagrange mean value theorem

, 20:12, 20 October 2011
no edit summary
* [[Increasing and differentiable implies nonnegative derivative]]
* [[Derivative of differentiable function on interval satisfies intermediate value property]]

==Facts used==

# [[uses::Continuous functions form a vector space]]
# [[uses::Differentiable functions form a vector space]]
# [[uses::Rolle's theorem]]
# [[uses::Differentiation is linear]]
==Proof==

{| class="sortable" border="1"
! 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) || <toggledisplay>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 $[a,b]$.</toggledisplay>
|-
| 4 || $g$ is differentiable on $(a,b)$ || Fact (2) || $f$ is differentiable on $(a,b)$ || Steps (1), (2) || <toggledisplay>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 $(a,b)$.</toggledisplay>
|-
| 5 || $g(a) = g(b) = 0$ || || || Steps (1), (2) || <toggledisplay>$h(a) = f(a)$ yields $g(a) = 0$. $h(b) = f(b)$ yields $g(b) = 0$.</toggledisplay>
|-
| 6 || There exists $c \in (a,b)$ such that $g'(c) = 0$. || Fact (3) || || Steps (3), (4), (5) || <toggledisplay>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.</toggledisplay>
|-
| 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
|}
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