# Second derivative test/con cavity

I don't feel like I'm understanding the actual concept.

I know that it's concave up when f''(x)>0 but I'm trying to visualise what happens to f(x) when the second derivative becomes positive or negative which I fail to. Sorry for this broad question, but it feels like I don't actually understand it apart from those basic conditions of when it's concave up or down.

Also, To clarify one thing, when the second derivative is zero at a stationary point (ie the x value where the first derivative is equal to zero), this tells us nothing useful right? This is different to when the value of a certain x which makes the second derivative equal zero, right? Because this would tell us the point of inflection?

(Also this is to a final year high school level math extent)

• @Olod so the first case is local minimum, right? Since the second derivative is positive? – Destudent Feb 19 '17 at 11:08
• Yes! If a function is twice differentiable at a critical point $c,$ and if $f''(c)$ is nonzero, then if $f''(c)>0$ the graph of $f(x)$ about $c$ will remind you a parabola opened UP with the vertex at $(c,f(c)),$ and if $f''(c)<0$ a parabola opened DOWN with the vertex at $(c,f(c)).$ So in the first case (convexity) we'll have a local minimum, and in the second case (concavity) we'll have a local maximum. – Olod Feb 19 '17 at 11:20

Consider a twice differentiable function $f$ on $\mathbb{R}$.
$$\begin{array}{c|c|c} & \color{blue}{f}\ \textrm{is decreasing} & \color{blue}{f}\ \textrm{is increasing}\\ & \implies f'(x) < 0 & \implies f'(x) > 0\\ \hline \color{orange}{f'}\ \textrm{is decreasing}\implies f''(x) > 0 & \textbf{concave up} & \textbf{concave up}\\ \hline \color{orange}{f'}\ \textrm{is increasing}\implies f''(x) < 0 & \textbf{concave down} & \textbf{concave down}\\ \end{array}$$