How can I understand the sign of second derivative at the extrema?

Question

Can someone explain why is the second derivative $\frac{d^2y}{dx^2}$ of a function $y=f(x)$ at a minimum (maximum) is positive (negative)?

Can we arrive at any conclusion from the definition of the second derivative as the derivative of the first derivative, and using the definition of derivative?

Answer 1

Suppose $x_0$ is a minimum of $f$, so that $f^\prime (x_0) = 0$.

Since $x_0$ is a minimum, $f$ must be increasing to the right of $x_0$. That is, $f^\prime$ must be positive in this neighbourhood. What does this imply about $f^{\prime \prime}(x_0)$?

Answer 2

My intuition is the following: At a minimum, a function $f$ goes from being decreasing to being increasing. In other words, its derivative $f'$ goes from being negative to being positive. So $f'$ is then increasing, i.e. $f''$ is positive.

Answer 3

Let $x_0$ be the critical point, that is: $$f'(x_0)=\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=0.$$ The second derivative test: Assume $f''(x_0)>0$, then: $$f''(x_0)=\lim_{x\to x_0} \frac{f'(x)-f'(x_0)}{x-x_0}=\lim_{x\to x_0} \frac{f'(x)-0}{x-x_0}>0.$$ Hence:

If $x\to x_0^-$ (i.e. $x$ approaches $x_0$ from the left side), then $x-x_0<0$ and $f'(x)<0$.

If $x\to x_0^+$ (i.e. $x$ approaches $x_0$ from the right side), then $x-x_0>0$ and $f'(x)>0$.

Hence, the function changes from a decrease ($f'(x)<0$) to an increase ($f'(x)>0$) over $x_0$, indicating the function achieves its local minimum at $x_0$.