If $f $ is defined on an open interval containing $x_0 $, if $f $ assumes its maximum or minimum at $x_0 $ and $f$ is differentiable at $x_0 $, then $f’(x_0) = 0$.
$f$ is continuous on $[ a, b ]$ and differentiable on $(a, b)$ and $f(a) = f(b)$
$$\exists x \in (a, b)\text{ s.t. }f’(x) = 0 $$
$f$ is continuous on $[ a, b ]$ and differentiable on $(a, b)$
$$\exists x \in (a, b)\text{ s.t. } f’(x) = \frac{ f(b) - f(a)}{ b - a } $$
Let $f$ be a differentiable function on $(a, b) $ such that $f’(x) = 0 $ for all $x \in (a, b)$. Then $f $ is a constant function on $(a, b)$.
Let $f$ be a real-valued function defined on an interval $I $. We say $f $ is strictly increasing on $I$ if
$x_1, x_2 \in I $ and $x_1 < x_2 $ imply $f(x_1) < f(x_2)$
$x_1, x_2 \in I $ and $x_1 > x_2 $ imply $f(x_1) > f(x_2)$
$x_1, x_2 \in I $ and $x_1 < x_2 $ imply $f(x_1) \leq f(x_2)$
$x_1, x_2 \in I $ and $x_1 > x_2 $ imply $f(x_1) \geq f(x_2)$
Let $f $ be a differentiable function on an interval $(a, b)$. Then 1. $f $ is strictly increasing if $f’(x) > 0 $ for all $x(a, b)$ 2. $f $ is strictly decreasing if $f’(x) < 0 $ for all $x(a, b)$ 3. $f $ is decreasing if $f’(x) \geq 0 $ for all $x(a, b)$ 4. $f $ is decreasing if $f’(x) \leq 0 $ for all $x(a, b)$
Let $f $ be a differentiable function on $(a, b) $. If $a< x_1 < x_2 < b $ and if $c$ lies between $f’(x_1) $ and $f’(x_2) $, there exists (at least one) $x $ in $(x_1, x_2) $ such that $f’(x) = c$.
Let $f $ be a one-to-one continuous function on an open interval $I $, and let $J = f(I) $. If $f $ is differentiable at $x_0 \in I$ and if $f’(x_0) \neq 0 $, then $\inv{ f } $ is differentiable at $y_0 = f(x_0) $ and
$$(\inv{ f })^\prime (y_0) = \frac{ 1 }{ f’(x_0)} $$