Let $I \subseteq \R$ be an interval and $f : I \to \R$ be a function. We denote the interior of $I$ by $I^\circ$.
Monotonicity
We say that $f$ is increasing if $f(x) \leq f(y)$ for all $x, y \in I$ with $x < y$ and that $f$ is strictly increasing if the former inequality is strict. We define decreasing and strictly decreasing analogously.
Characterization of monotonicity
Suppose that $f$ is continuous on $I$ and differentiable on $I^\circ$.1 Then $f$ is increasing on $I$ if and only if $f’ \geq 0$ on $I^\circ$. In addition, $f$ is strictly increasing on $I$ if $f’ > 0$ on $I^\circ$.2
Extremality
We say that $x \in I$ is a (global) minimizer and that the value $f(x)$ is a (global) minimum of $f$ if $f(x) \leq f(y)$ for all $y \in I$. We define (global) maximizer and (global) maximum analogously.
Extreme value theorem
If $I$ is closed and bounded and $f$ is continuous on $I$, then $f$ has a minimizer and a maximizer in $I$.
We say that $x \in I$ is a local minimizer and that the value $f(x)$ is a local minimum of $f$ if there exists an $\epsilon > 0$ such that $f(x) \leq f(y)$ for all $y \in I \cap (x-\epsilon, x+\epsilon)$. We define local maximizer and local maximum analogously. Clearly, any global extremizer must also be a local extremizer.
Fermat’s theorem
Suppose that $x \in I^\circ$ is a local extremizer of $f$. If $f$ is differentiable at $x$, then $f’(x) = 0$ (that is, $x$ is a stationary point of $f$).
Thus, under the hypotheses of the extreme value theorem, any extremizer of $f$ will either be an interior point of $I$ and hence a point at which $f$ is stationary or not differentiable, or else will be a boundary point of $I$. We call a point of the former type (that is, a point at which $f$ is stationary or not differentiable) a critical point of $f$.
First derivative test
If $f$ is continuous at $x$ and there exists an $\epsilon > 0$ such that $f’ \leq 0$ on $(x-\epsilon, x)$ and $f’ \geq 0$ on $(x, x+\epsilon)$, then $x$ is a local minimizer of $f$.
Second derivative test
If $f$ is stationary at $x$ and $f''(x) > 0$, then $x$ is a local minimizer of $f$.
Convexity
We say that $f$ is convex if $f((1-\theta) x + \theta y) \leq (1-\theta)f(x) + \theta f(y)$ for all distinct $x, y \in I$ and $\theta \in (0, 1)$ and that $f$ is strictly convex if the inequality is strict. We define concave and strictly concave analogously.
Secant and tangent line characterizations of convexity
Given an $x \in I$, let $g(y; x) := \frac{f(y)-f(x)}{y-x}$ for all $y \in I \setminus \set{x}$. Then $f$ is convex (resp., strictly convex) on $I$ if and only if $g$ is increasing (resp., strictly increasing) in $y$ for all $x \in I$. In addition, if $f$ is differentiable on $I^\circ$, then $f$ is convex (resp., strictly convex) on $I$ if and only if $g(y; x) \geq f'(x)$ (resp., $g(y; x) > f'(x)$) for all $x \in I^\circ$ and $y \in I \setminus \set{x}$.
First-order characterization of convexity
Suppose that $f$ is continuous on $I$ and differentiable on $I^\circ$. Then $f$ is convex (resp., strictly convex) on $I$ if and only if $f’$ is increasing (resp., strictly increasing) on $I^\circ$.
Second-order characterization of convexity
Suppose that $f$ is continuous on $I$ and twice differentiable on $I^\circ$. Then $f$ is convex on $I$ if and only if $f'' \geq 0$ on $I^\circ$. In addition, $f$ is strictly convex on $I$ if $f'' > 0$ on $I^\circ$.
Minimizers of convex functions
If $f$ is convex and $x$ is a local minimizer of $f$, then $x$ is a global minimizer of $f$. In addition, if $f$ is strictly convex, then $f$ has at most one local minimizer.
We say that the graph of $f$ has an inflection point at $(x, f(x))$ if it has a tangent line at $(x, f(x))$ (which may be vertical) and its (signed) curvature $\kappa := \frac{f''}{[1+(f')^2]^{3/2}}$ changes sign at $x$ (though $\kappa$ need not exist at $x$ itself). At such a point, $f$ changes from strictly convex to strictly concave or vice-versa.