Maxima & minima formulas

Necessary condition: if $f$ has a local extremum at an interior point $c$ and $f^{\prime }(c)$ exists, then $f^{\prime }(c)=0$. So every interior local extremum is a critical point. The converse is false: $f(x)=x^3$ has $f^{\prime }(0)=0$ but no extremum at $0$.

$I$ is an interval, $f$ is continuous on $I$ and differentiable at its interior points. $f^{\prime }\ge 0$ on $I$ gives non-decreasing; if in addition $f^{\prime }=0$ only at isolated points, $f$ is still strictly increasing. The converse of the first implication fails: $x^3$ is strictly increasing on $\mathbb{R}$ although $f^{\prime }(0)=0$.

Needs $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)$ to exist, so it cannot be used at a critical point where $f^{\prime }(c)$ does not exist. If $f^{\prime \prime }(c)=0$ the test is inconclusive and you must use the first-derivative test or higher derivatives.

Both conditions matter: continuity, and a closed bounded interval. On an open interval an extremum may not exist, for example $f(x)=x$ on $(0,1)$ has neither. A discontinuous function on a closed interval may also fail, for example $f(x)=x$ for $0\le x<1$ with $f(1)=0$ has no maximum on $[0,1]$.

Convex (also called concave up) means $f(\lambda x+(1-\lambda )y)\le \lambda f(x)+(1-\lambda )f(y)$ for all $x,y\in I$ and $\lambda \in [0,1]$. For differentiable $f$ this is equivalent to the graph lying on or above every tangent line: $f(y)\ge f(x)+f^{\prime }(x)(y-x)$ for all $x,y\in I$. $f^{\prime \prime }>0$ on $I$ makes $f$ strictly convex, but not conversely: $x^4$ is strictly convex although $f^{\prime \prime }(0)=0$.

A point of inflection is a point $c$ where $f$ is continuous and the graph changes from convex to concave or the reverse. If $f^{\prime \prime }(c)$ exists at an inflection point then $f^{\prime \prime }(c)=0$. The converse is false: $f(x)=x^4$ has $f^{\prime \prime }(0)=0$ but $f^{\prime \prime }\ge 0$ on both sides, so $0$ is not an inflection point. An inflection point can also occur where $f^{\prime \prime }$ does not exist: $f(x)=x|x|$ has $f^{\prime \prime }(x)=-2$ for $x<0$ and $2$ for $x>0$, with $f^{\prime \prime }(0)$ undefined.

The same holds with min in place of max. Reason: if some $x_1\in I$ had $f(x_1)>f(c)$, $f$ would have a local minimum strictly between $c$ and $x_1$, which would be a second critical point.

The iterate stays put exactly when $f^{\prime }(x_k)=0$, that is at a critical point. That point need not be a minimum if the iterate lands on it exactly. On a non-convex $f$ the iteration can settle at a local minimum that is not global; which one depends on $x_0$ and $\eta$.

Converges to $x^{\ast }$ from every starting point if and only if $|1-\eta a|<1$, that is $0<\eta <2/a$. With $\eta =1/a$ it reaches $x^{\ast }$ in one step. For $1/a<\eta <2/a$ the iterates alternate sides of $x^{\ast }$ while converging. At $\eta =2/a$ they bounce between $x_0$ and $2x^{\ast }-x_0$ forever, and for $\eta >2/a$ they diverge (whenever $x_0\ne x^{\ast }$). Here $a=f^{\prime \prime }$, the constant curvature.