What is the meaning of convexity for a function on an interval?

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Convexity of a function on an interval means that the function's graph lies below the straight lines connecting any two points on the graph within that interval. A function is convex at a point if its second derivative is positive, while it is concave if the second derivative is negative. This concept can also apply to non-differentiable functions, where a function is considered convex if the set of points above its graph forms a convex set. For example, the parabola x^2 is convex, while -x^2 is concave. The discussion confirms that the function x^x is indeed convex for x > 0.
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What does it mean for a function to be convex (or concave) on an interval [a,b]? I understand what a function is and what an interval is, but I don't get what "convexity" is.
 
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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, ifp = (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.
 
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A convex set is a set where all points can be connected with a straight line inside the set (so every point can "see" every other). A function is convex if the set above it (ie the set {(x,y):y>f(x)}) is convex.

If the function is twice differentiable, this is equivalent with that the second derivative is everywhere non-negative.
 
Data said:
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, ifp = (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?
 
Yes. (damn character limit)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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