Search results
Results from the WOW.Com Content Network
A differentiable function f is (strictly) concave on an interval if and only if its derivative function f ′ is (strictly) monotonically decreasing on that interval, that is, a concave function has a non-increasing (decreasing) slope. [3] [4] Points where concavity changes (between concave and convex) are inflection points. [5]
The second derivative of a function f can be used to determine the concavity of the graph of f. [2] A function whose second derivative is positive is said to be concave up (also referred to as convex), meaning that the tangent line near the point where it touches the function will lie below the graph of the function.
Every concave function that is nonnegative on its domain is log-concave. However, the reverse does not necessarily hold. An example is the Gaussian function f(x) = exp(−x 2 /2) which is log-concave since log f(x) = −x 2 /2 is a concave function of x. But f is not concave since the second derivative is positive for | x | > 1:
Specifically, a twice-differentiable function f is concave up if ″ > and concave down if ″ <. Note that if f ( x ) = x 4 {\displaystyle f(x)=x^{4}} , then x = 0 {\displaystyle x=0} has zero second derivative, yet is not an inflection point, so the second derivative alone does not give enough information to determine whether a given point is ...
A sigmoid function is a bounded, differentiable, real function that is defined for all real input values and has a non-negative derivative at each point [1] [2] and exactly one inflection point. Properties
In simple terms, a convex function graph is shaped like a cup (or a straight line like a linear function), while a concave function's graph is shaped like a cap . A twice-differentiable function of a single variable is convex if and only if its second derivative is nonnegative on its entire domain. [1]
As can be seen, this function is not convex because of the concavity, and it is not pseudoconvex because it is not differentiable at =. Quasiconvex function that is not convex, nor pseudoconvex. Generalization to nondifferentiable functions
If one wanted to solve the problem with standard tools such as the implicit function theorem, one would have to assume that the problem is well behaved: U(.) is twice continuously differentiable, concave in s, that the domain over which s is defined is convex, and that it there is a unique maximizer () for every value of p and that () is in the ...