$f(x)=x^3$ is convex nor concave without to derivate

Question

Drawing the graph of a function is not intuitive. For example, the function $f(x)=x^2$ is special, we can say that given two points of the function the straight line that join those points is above the function. So we write $$ x_2<(x_1+x_0)(x-x_0)+x^2 $$ How did I obtain the part of the right? $$ (Y-x_0^2)/ (x-x_0) = (x_1^2 - x_0^2) / (x_1 - x_0) $$ Now I just isolate Y in the equation above.

My question is, can I do the same with $f(x)=x^3$? And how do I prove it? Can I say that $x^3$ is below the straight line that join two points of the function $x^3$? If and only if $x$ is positive.

Answer 1

For a graph of a function to be convex the epigraph must be a convex set. This means I can take any two points in the set and draw a straight line between them without leaving the set. To see that this isn't true for $f(x)=x^3$ take any $x < 0$ and draw the line from $(x,f(x))$ to the point $(0,0)$ which are both in the epigraph of the function, but none of the points between them are in the epigraph so $f(x)$ cannot be convex. The concave case is the same but with $x>0$.

Answer 2

Hint:   for $\,x_{0,1} \ge 0\,$ the generalized means inequality gives $\,\sqrt[3]{\dfrac{x_0^3+x_1^3}{2}} \ge \dfrac{x_0+x_1}{2}\,$, which is equivalent to $\,f(x)=x^3\,$ being midpoint-convex on $\,\Bbb R^+\,$. By continuity, it follows that $\,f(x)\,$ is in fact convex on $\,\Bbb R^+\,$. More generally, the same argument works for $\,f(x)=x^n\,$, $\,n \ge 1\,$.

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[ EDIT ]   Since $\,f(x)=x^3\,$ is an odd function, it follows that it is concave on $\,\Bbb R^-\,$. Since the direction of concavity changes at $0$, the function is neither convex nor concave on the entire $\,\Bbb R^\,$.