0
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ @Olod so the first case is local minimum, right? Since the second derivative is positive? $\endgroup$ – Destudent Feb 19 '17 at 11:08
  • $\begingroup$ 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. $\endgroup$ – Olod Feb 19 '17 at 11:20
1
$\begingroup$

Consider a twice differentiable function $f$ on $\mathbb{R}$.

$$\begin{array}{c|c|c} & \color{blue}{f}\ \textrm{is decreasing} & \color{blue}{f}\ \textrm{is increasing}\\ & \implies f'(x) < 0 & \implies f'(x) > 0\\ \hline \color{orange}{f'}\ \textrm{is decreasing}\implies f''(x) > 0 & \textbf{concave up} & \textbf{concave up}\\ \hline \color{orange}{f'}\ \textrm{is increasing}\implies f''(x) < 0 & \textbf{concave down} & \textbf{concave down}\\ \end{array}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.