7.3 Second derivative - concavity

There are two intervals where the function is concave up, and two where it is concave down.

In the graph above, the concave-up sections are marked in red. These are where the graph has an increasing derivative; equivalently, where the graph lies above its tangent lines; more descriptively, where it curves like a smiley face.

Concave-down sections are marked in blue. These are where the graph has a decreasing derivative; equivalently, where the graph lies below its tangent lines; more descriptively, where it curves like a frowney face.

Try allowing your graph to have horizontal asymptotes. For example, let the function get closer and closer to the $x$-axis (or another horizontal line) without touching it.

The most basic shape of the graph is given by the last two bullet points:

The curve is concave down over the interval $(-5,5)$, so let's give it a frowney-face curvature there.

Finally, when $x>5$ or $x<-5$, our curve should be concave up, so let's give it smiley-face curvature there, without changing its basic increasing/decreasing shape.

This finishes our sketch.

Consider $f(x)=(x-3)^4$.

In general, false.

An inflection point is where the concavity of a function changes. It is possible that $x=3$ is an inflection point, but it is also possible that is not. So, the statement is false, in general.

For example, let $f(x)=(x-3)^4$. Since $f(x)$ is a polynomial, all its derivatives exist and are continuous. $f''(x)=12(x-3)^2$, so $f''(3)=0$. However, since $f''(x)$ is something squared, it is never negative, so $f(x)$ is never concave down. Since $f(x)$ is never concave down, it never changes concavity, so it has no inflection points.

Remark: finding inflection points is somewhat reminiscent of finding local extrema. To find local extrema, we first find all critical and singular points, since local extrema can only occur there or at endpoints. Then, we have to figure out which critical and singular points are actually local extrema. Similarly, if you want to find inflection points, start by finding where $f''(x)$ is zero or non-existant, because inflection points can only occur there. Then, you still have to check whether those points are actually inflection points.

$x=1,\ y=11$

Inflection points occur where $f''(x)$ changes sign. Since $f(x)$ is a polynomial, its first and second derivatives exist everywhere, and are themselves polynomials. In particular,

The second derivative is negative for $x<1$ and positive for $x>1$. Thus the concavity changes between concave up and concave down at $x=1,\ y=11$.

This is the only inflection point. It is true that $f''(0)=0$, but for values of $x$ both a little larger than and a little smaller than 0, $f''(x)<0$, so the concavity does not change at $x=0$.