# Convexity of a Function

by Izzhov
Tags: convexity, function
 P: 998 A (twice-differentiable) function is concave at a point if its second derivative is negative at that point. Similarly a function is convex at a point if its second derivative is positive at that point. You can extend the definition to functions that aren't differentiable also; see http://en.wikipedia.org/wiki/Concave_function. Intuitively: A concave (or "concave down") function is one that is "cupped" downwards. For example, the parabola $-x^2$ is concave throughout its domain, and the parabola $x^2$ is convex throughout its domain. There are functions which are "cupped" but don't actually have the cup shape. For example, $1/x$ is concave on the negative reals and convex on the positive reals, however it doesn't have any extrema at all. Another way to present it is: A function $f$ is convex on an interval if the set of points above its graph on that interval is a convex set; that is, if$p = (x_1, y_1)$ and $q = (x_2, y_2)$ are points with $x_1, x_2$ on the interval of interest, $y_1 \geq f(x_1)$, and $y_2 \geq f(x_2)$, then the straight line joining $p$ to $q$ lies entirely above the graph of $f$. Then you can define $f$ is concave whenever $-f$ is convex.
 Quote by Data A (twice-differentiable) function is concave at a point if its second derivative is negative at that point. Similarly a function is convex at a point if its second derivative is positive at that point. You can extend the definition to functions that aren't differentiable also; see http://en.wikipedia.org/wiki/Concave_function. Intuitively: A concave (or "concave down") function is one that is "cupped" downwards. For example, the parabola $-x^2$ is concave throughout its domain, and the parabola $x^2$ is convex throughout its domain. There are functions which are "cupped" but don't actually have the cup shape. For example, $1/x$ is concave on the negative reals and convex on the positive reals, however it doesn't have any extrema at all. Another way to present it is: A function $f$ is convex on an interval if the set of points above its graph on that interval is a convex set; that is, if$p = (x_1, y_1)$ and $q = (x_2, y_2)$ are points with $x_1, x_2$ on the interval of interest, $y_1 \geq f(x_1)$, and $y_2 \geq f(x_2)$, then the straight line joining $p$ to $q$ lies entirely above the graph of $f$. Then you can define $f$ is concave whenever $-f$ is convex.
So this means the function $$x^x$$ is convex where x>0, correct?