# How zeros of second derivative map to equilibrium value

This question arose from plotting some functions in MATLAB and GeoGebra. Assume we have a function of the form

$$f(x)=A \sin(P_n(x))+d \qquad \text{or}\qquad f(x)=A \cos(P_n(x))+d$$

where

$$P_n(x) = \sum_{i=0}^n a_i x^i,\quad a_i \in \mathbb{R}.$$

How do we show that if $f''(x)=0$, then $f(x)=A/2+d$?

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Well, you can't ... take $A = 1$, $d = 0$, and $P_n(x) = x^2$. Then if $f(x) = \cos(x^2)$, $f''(x) = -2\sin(x^2) - 4x^2\cos(x^2)$. This has $f''(0) = 0$ but $f(0) = 1\neq \frac{1}{2}$.
Indeed!, but it seems to be correct if one looks at the interval $x>0$, no? =) –  N3buchadnezzar Oct 26 '12 at 21:38
Translate right: $\cos((x-1)^2)$. –  Neal Oct 26 '12 at 21:43