Convexity and concavity of a function

[y_lindsay]

here is the problem:
we know that
(1) f(x) is defined on $$(-\infty, +\infty)$$, and f(x) has the second derivative everywhere,
(2) $$lim_{|x|\rightarrow+\infty}(f(x)-|x|)=0$$,
(3) there is $$x_0\in R$$ such that $$f(x_0)\leq0$$
how do we prove that f"(x) changes sign on $$(-\infty, +\infty)$$?

I can imagine that the graph of f(x) can't be always convex or concave on the real line, but how do I put it in hard argument?

 

[birulami]

Funny you did not get an answer yet, as I consider this an interesting problem. The conclusion seems to be so obvious when you try to draw the graph.

 

[tacman]

Convexity and concavity of a function are important concepts in calculus that describe the shape of a function's graph. A function is said to be convex if its graph curves upward, and concave if its graph curves downward. In this problem, we are given a function f(x) that satisfies certain conditions and we need to prove that its second derivative, f"(x), changes sign on the entire real line.

Firstly, let's consider the given condition (2) that states lim_{|x|\rightarrow+\infty}(f(x)-|x|)=0. This means that as x approaches positive or negative infinity, the function f(x) approaches the line y = |x|. This line is a straight line with a slope of 1, which is neither convex nor concave. Therefore, we can conclude that the function f(x) cannot be always convex or concave on the real line.

Next, we will use the given condition (3) that states there is x_0\in R such that f(x_0)\leq0. This means that there exists a point x_0 on the x-axis where the function f(x) is either equal to or below the x-axis. This point is known as a point of inflection, where the concavity of the function changes. Since the function f(x) is continuous and has a second derivative everywhere, we can apply the Second Derivative Test to determine the concavity at x_0. If f"(x_0) > 0, then the function is concave at x_0, and if f"(x_0) < 0, then the function is convex at x_0.

Now, we can use the Intermediate Value Theorem to show that there must be a point where f"(x) changes sign on the real line. Since f"(x) is continuous on the entire real line, it must take on all values between f"(x_0) and f"(+\infty) as x varies from x_0 to +\infty. Similarly, it must take on all values between f"(x_0) and f"(-\infty) as x varies from x_0 to -\infty. Therefore, there must be a point between x_0 and +\infty where f"(x) changes from positive to negative, and another point between x_0 and -\infty where f"(