PlanetPhysics/Inflexion Point

In examining the [[../Cod/|graphs]] of differentiable real [[../Bijective/|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 $f^{\prime }$ 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, $f^{\prime }$ 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\, $x=0,\pm \pi ,\pm 2\pi ,\dots$;\, $f^{‴}(x)=-\cos x$, $f^{‴}(n\pi )=-\cos n\pi =(-1{)}^{n+1}\ne 0$. 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 $\kappa (x)=\frac{f^{″}(x)}{[1+f^{\prime }(x{)}^2{]}^{3/2}},$ 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 \,$f^{\prime }(\xi )=0$,\, the point is said to be a stationary inflexion point or a saddle-point , while in the case\, $f^{\prime }(\xi )\ne 0$\, it is a non-stationary inflexion point .