You cannot. The function f(x) = x2 + 1 has no real zeros. But it does have a minimum.
Given an undirected graph G=(V,E) and an integer k, find induced subgraph H=(U,F) of G of maximum size (maximum in terms of the number of vertices) such that all vertices of H have degree at least k
Let G be a complete graph with n vertices. Consider the case where n=2. With only 2 vertices it is clear that there will only be one edge. Now add one more vertex to get n = 3. We must now add edges between the two old vertices and the new one for a total of 3 vertices. We see that adding a vertex to a graph with n vertices gives us n more edges. We get the following sequence Edges on a graph with n vertices: 0+1+2+3+4+5+...+n-1. Adding this to itself and dividing by two yields the following formula for the number of edges on a complete graph with n vertices: n(n-1)/2.
In a connected component of a graph with Mi vertices, the maximum number of edges is MiC2 or Mi(Mi-1)/2. So if we have k components and each component has Mi vertices then the maximum number of edges for the graph is M1C2+M2C2+...+MKC2. Of course the sum of Mi as i goes from 1 to k must be n since the sum of the vertices in each component is the sum of all the vertices in the graph which you gave as n. Where MC2 means choose 2 from M and there are M(M-1)/2 ways to do that.
A star graph, call it S_k is a complete bipartite graph with one vertex in the center and k vertices around the leaves. To be a tree a graph on n vertices must be connected and have n-1 edges. We could also say it is connected and has no cycles. Now a star graph, say S_4 has 3 edges and 4 vertices and is clearly connected. It is a tree. This would be true for any S_k since they all have k vertices and k-1 edges. And Now think of K_1,k as a complete bipartite graph. We have one internal vertex and k vertices around the leaves. This gives us k+1 vertices and k edges total so it is a tree. So one way is clear. Now we would need to show that any bipartite graph other than S_1,k cannot be a tree. If we look at K_2,k which is a bipartite graph with 2 vertices on one side and k on the other,can this be a tree?
I believe a cubic rectangle has four vertices
cubic function cubic function
The inverse of the cubic function is the cube root function.
In general, there is no simple method.
YES!!!! However, y = x^(3) does not have a vertex , but an horizontal point at the origin. However, y = ax^(3) + bx^(2) + dx + c will have vertices , and they are found by differemntiating, and equating to zero.
it is a vertices's form of a function known as Quadratic
A cubic graph!
A cubic graph must have an even number of vertices. Then, a Hamilton cycle (visiting all vertices) must have an even number of vertices and also an even number of edges. Alternatively color this edges red and blue, and the remaining edges green.
No, an function only contains a certain amount of vertices; leaving a logarithmic function to NOT be the inverse of an exponential function.
It is the sum of the y-coordinates of the vertices divided by the number of vertices.
The cubic function.
No. For a volume you must have a cubic function.