Localização relativa da concavidade da função

A tangent line to a curve lies above the curve if it is concave down, and it lies below the curve if it is concave up.

Here, let us examine a function $f(x)$ that is concave down at $x=a$. Thus, $f''(a)<0$.

Also, we know that $f''(a+\epsilon)<0$ and $f''(a-\epsilon)<0$ for sufficiently small $\epsilon>0$.

Thus, we have that $f'(a-\epsilon)>f'(a)>f'(a+\epsilon)$.

Suppose that the linear approximation for $f(x)$ at $x=a$ is $y=g(x)$.

Then, for all real $x$, $g'(x)=f'(a)$, since $g$ is tangent to $f$ at $x=a$.

Thus, for all $x$ in $(a-\epsilon,a)$, we have that $g'(x)<f'(x)$.

Similarly, for all $x$ in $(a,a+\epsilon)$, we have that $g'(x)>f'(x)$.

The two inequalities above lead us to the following two inequalities in integrals: $$\int_{a-\epsilon}^{a}g'(x)dx<\int_{a-\epsilon}^{a}f'(x)dx$$ $$\int_{a}^{a+\epsilon}g'(x)dx>\int_{a}^{a+\epsilon}f'(x)dx$$ Simplifying the two inequalities above, $$g(a)-g(a-\epsilon)<f(a)-f(a-\epsilon)$$ $$g(a+\epsilon)-g(a)>f(a+\epsilon)-f(a)$$ Finally, since $g$ is a linear approximation at $x=a$, $g(a)=f(a)$, which leads us to: $$g(a-\epsilon)>f(a-\epsilon)$$ $$g(a+\epsilon)>f(a+\epsilon)$$ Thus, for sufficiently small deviations from the point of tangency, the tangent lies above the curve it approximates if the curve is concave down. This is exactly what we wanted to prove.

Using a nearly identical proof, we may say that the tangent lies below the curve if it is concave up.

Last, if we take the tangent line at a point of inflection, then we may apply the argument above, but only to one side of the curve. Thus, on the side that is concave down, the line is above the curve, and it is the other way around on the side that is concave up.