Prove that if a function $f(x)$ is concave down on the interval $(a,b)$ then the second derivative cannot be greater than $0$ for $x \in (a, b)$ I have tried approaching this by drawing out the graph and the second derivative but I’m thoroughly confused on where to start and what would constitute a proof of this. I believe I am supposed to use some form of MVT but I’m not sure where that would fit into this.
 A: Here is an outline for the proof.
Note that for $x$ in a neighborhood of $a$, $$f(x)=f(a)+f'(a)(x-a)+\frac {f''(\zeta)}{2}(x-a)^2$$
The first part of the RHS is the equation of the tangent line.
For the function to be concave down you have to have your $f(x)$ to be below the tangent line so $$  \frac {f''(\zeta)}{2}(x-a)^2\le 0$$
Thus your second derivative should be less than or equal zero.
A: We can't use Taylor's theorem with mean-value form of the remainder in the general case, because it would need the continuity of the second derivative. Instead, we can use Taylor's Theorem with Peano's Form of Remainder. Let $c \in (a,b)$ be given. We have that
$$f(x)=f(c)+f'(c)(x-c)+\frac{1}{2}f''(c)(x-c)^2+o((x-c)^2)$$
Concavity implies that the tangent line is over the function, i.e
$$f(x)\leqslant f(c)+f'(c)(x-c)$$
Which implies that
$$\frac{1}{2}f''(c)(x-c)^2+o((x-c)^2)\leqslant 0$$
I.e. for $x\neq c$ we have that
$$\frac{1}{2}f''(c)+o(1)\leqslant 0$$
And this implies that the derivative is negative, because if we take $x \to c$ we get that
$$\frac{1}{2}f''(c)\leqslant 0$$
