I was reading a book on Calculus, by Michael Spivak. There they mention that points where the derivative is equal to zero are called critical points. They nowhere mention that where the derivative does not exist will be a critical point. When I Googled it, they say critical points are those where the function derivative is either zero or the derivative does not exist. Which definition should I take?
[Math] the definition of a Critical Point
calculus
Related Solutions
The point $(0,0)$ is a minimum point. It is also an undulation point. You are right that in some ways this is a poor example of an undulation point, since it also has other properties. On the other hand, this example does make the point easy to see, and it has an extremely easy formula.
A better example in some ways would be $(0,0)$ in the graph of $f(x)=x^4+x$. The point is still fairly noticeable but is not a minimum.
ADDED: I just took at look at the Wikipedia definition of undulation point, and it does give the example $f(x)=x^4$ in the text. However, it also has the example $y=x^4-x$ in a graph later in the article. This is the same as my example but reflected in the $y$ axis. It seems Wikipedia wanted to have it both ways.
As you say, there are no critical points, so you can conclude that f has no relative extrema.
Furthermore, since $\displaystyle f^{\prime}(x)=-\frac{x^2+1}{(x^2-1)^2}<0$ for $x\ne\pm1$, you can also conclude that
$\hspace{.4 in}f$ is decreasing on the intervals $(-\infty, -1),\;(-1,1),\;\text{and }(1,\infty)$.
Since the second derivative $f^{\prime\prime}$ equals 0 at $x=0$ and is undefined for $x=\pm1$,
you have to determine the sign of $f^{\prime\prime}$ on each of the intervals $(-\infty,-1), (-1,0), (0,1), \text{and }(1,\infty)$
to determine the concavity of the graph.
Best Answer
When the derivative is 0 at a point $(x,y)$, that point is critical. When a derivative does not exist, there might be no single point that can be labeled as critical. For example, the function $x, x\in (-\infty, 0)$ and $x+3, x\in [0, \infty)$. The derivative does not exist at $x=0$, however there is no single point that can be labeled as critical. So in my opinion, if $\lim_{x\to a} f(x) $ exists, then that is a critical point.