PlanetPhysics/Inflexion Point

In examining the graphs of differentiable real functions, it may be useful to state the intervals where the function is convex and the ones where it is concave.


 * A function $$f$$ is said to be convex on an interval if the restriction of $$f$$ to this interval is a (strictly) convex function; this may be characterized more illustratively by saying that the graph of $$f$$ is concave upwards  or concave up . On such an interval, the tangent line of the graph is constantly turning counterclockwise, i.e., 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\, $$(-\pi,\,0)$$.
 * The concavity of the function $$f$$ on an interval correspondingly: On such an interval, the graph of $$f$$ is concave downwards  or 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\, $$(0,\,\pi)$$.
 * The points in which a function changes from concave to convex or vice versa are the inflexion points (or 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.

\begin{pspicture}(-5,-2.5)(5,2) \psaxes[Dx=9,Dy=1]{->}(0,0)(-4.5,-1.5)(5,2) \rput(5,-0.2){$$x$$} \rput(0.2,2){$$y$$} \rput(3,-0.2){$$\pi$$} \rput(-3.1,-0.2){$$-\pi$$} \psplot[linecolor=blue]{-4}{4}{x 60 mul sin} \psdot[linecolor=red](0,0) \rput(0.2,-2.3){The origin is an inflexion point of the sinusoid \,$$y = \sin{x}$$.} \end{pspicture}

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.

Remarks

1. For finding the inflexion points of the graph of $$f$$ it does not suffice to find the roots of the equation\, $$f(x) = 0$$, since the sign of $$f$$ does not necessarily change as one passes such a root. If the second derivative maintains its sign when one of its zeros is passed, we can speak of a 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 curvature $$\kappa$$ 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 stationary inflexion point or a saddle-point, while in the case\, $$f'(\xi) \neq 0$$\, it is a non-stationary inflexion point.