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Mean Value Theorem:  Suppose \hfil \break
1.) $f$ continuous on $[a, b]$
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2.) $f$ differentiable on $(a, b)$
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Then there exists $c \in (a, b)$ such that $f'(c) = {f(b) - f(a) \over b - a}$

Ex 3:  If $f'(x) = 0$ for all $x \in (a, b)$, then $f(x) = c$ for some constant $c$.

Proof:  Take $x_0 \in (a, b)$

Show $f(x) = f(x_0)$ for all $x \in (a, b)$ [i.e., $c = f(x_0)$].

Without loss of generality, assume $x > x_0$. \hfil \break
(proof is similar when $x < x_0$).

$f$ is continuous on $[x_0, x]$. \hfil \break
$f$ is differentiable on $(x_0, x).$

By MVT, there exists $c \in (x_0, x)$ such that 

${f(x) - f(x_0) \over x - x_0} = f'(c) = 0.$

Thus $f(x) - f(x_0) = 0$.  Hence $f(x) = f(x_0).$


Ex 4:  If $f'(x) = g'(x)$ for all $x \in (a, b)$, then $f(x) = g(x) + c$ for some constant $c$.

Proof:  $f'(x) = g'(x)$ implies $(f - g)'(x) = f'(x) - g'(x) = 0.$


Thus $(f - g)(x) = f(x) - g(x) = c$ for some constant $c$.  \hfil \break
Thus  $f(x) = g(x) + c.$ 

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The Fundamental Theorem of Calculus:  Suppose $f$ continuous on $[a, b]$.

1.)  If $g(x) = \int_a^x f(t) dt$, then $g'(x) = f(x)$.

2.)   $\int_a^b f(t) dt = F(b) - F(a)$ where $F$ is any antiderivative of $f$, that is $F' = f$.


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