inflexion pointInflexionPoint2007-10-01 16:36:172009-08-04 12:53:48Definitionconvex functionconcave upwardsconcave upconcave downwardsconcave downsaddle-pointplain point (?)second derivativecurvature% this is the default PlanetMath preamble. as your knowledge
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In examining the graphs of differentiable real functions, it may be useful to \PMlinkescapetext{state} the intervals where the function is \PMlinkescapetext{convex} and the ones where it is \PMlinkescapetext{concave}.
\begin{itemize}
\item A function $f$ is said to be \PMlinkescapetext{{\em convex} on an interval} if the \PMlinkname{restriction}{RestrictionOfAFunction} of $f$ to this interval is a (strictly) convex function; this may be characterized more illustratively by saying that the graph of $f$ is \PMlinkescapetext{{\em concave upwards}} or \PMlinkescapetext{\emph{concave up}}. On such an interval, the tangent line of the graph is constantly turning counterclockwise, \PMlinkname{i.e.}{Ie}, the derivative $f'$ is increasing and thus the second derivative $f''$ is positive. In the picture below, the sine curve is concave up on the interval\, $(-\pi,\,0)$.
\item The {\em concavity} of the function $f$ on an interval correspondingly: On such an interval, the graph of $f$ is \PMlinkescapetext{{\em concave downwards}} or \PMlinkescapetext{\emph{concave down}}, the tangent line turns clockwise, $f'$ decreases, and $f''$ is negative. In the picture below, the sine curve is concave down on the interval\, $(0,\,\pi)$.
\item The points in which a function changes from \PMlinkescapetext{concave to convex} or vice versa are the {\em inflexion points} (or \emph{inflection points}) of the graph of the function. At an inflexion point, the tangent line crosses the curve, the second derivative vanishes and changes its sign when one passes through the point.
\end{itemize}
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Since the sine function is $2\pi$-periodic, the sinusoid possesses infinitely many inflexion points. Indeed,\,
$f(x) = \sin x$;\, $f''(x) = -\sin x = 0$\, for\, $x = 0,\,\pm\pi,\,\pm2\pi,\,\dots$;\, $f'''(x) = -\cos x$, $f'''(n\pi) = -\cos n\pi = (-1)^{n+1} \neq 0$. Non-nullity of the third derivative at these critical points assures us the existence of those inflexion points.
\textbf{Remarks}
1. For finding the inflexion points of the graph of $f$ it does not suffice to find the \PMlinkname{roots}{Equation} of the equation\, $f''(x) = 0$, since the sign of $f''$ does not necessarily change as one passes such a \PMlinkescapetext{root}. If the second derivative maintains its sign when one of its zeros is passed, we can speak of a {\em plain point} (?) of the graph. E.g. the origin is a plain point of the graph of\, $x\mapsto x^4$.
2. Recalling that the \PMlinkname{curvature}{CurvaturePlaneCurve} $\kappa$ for a plane curve \,$y = f(x)$\, is given by
$$\kappa(x) \;=\; \frac{f''(x)}{[1+f'(x)^2]^{3/2}},$$
we can say that the inflexion points are the points of the curve where the curvature changes its sign and where the curvature equals zero.
3. If an inflexion point\, $x = \xi$\, satisfies the additional condition \,$f'(\xi) = 0$,\, the point is said to be a \PMlinkescapetext{{\em stationary inflexion point}} or a {\em saddle-point}, while in the case\, $f'(\xi) \neq 0$\, it is a {\em non-stationary inflexion point}.