How to prove a function is concave? (Single Variable)

Question

It has been a while after completing the calculus of single variable.

Right now I have a function of single variable $f(x)$, and that $f'(x)=-c$ for all $x$. So $f$ is a decreasing function.

Bu, $f''(x)=0$. Can I say that $f$ is a concave function?

Before posting this question, I did some research and consult with https://www.economics.utoronto.ca/osborne/MathTutorial/CV1F.HTM. Quoting from this website, it says that:

Proposition A twice-differentiable function f of a single variable defined on the interval I is concave if and only if f ''(x) ≤ 0 for all x in the interior of I convex if and only if f ''(x) ≥ 0 for all x in the interior of I.

Answer 1

Your function graph is a straight line. If you consider this to be concave, it is a trivial case.

Answer 2

From earlier on that same page:

Definition Let $f$ be a function of a single variable defined on an interval. Then $f$ is

• concave if every line segment joining two points on its graph is never above the graph
• convex if every line segment joining two points on its graph is never below the graph

We have $f(x) = -cx + k$, for some $c$, $k$. Every line segment joining two points on this graph is on the graph - neither above nor below. So $f$ is both concave and convex.