Consider the function $f$ represented below: enter image description here

Find the x-coordinates of the points at which the function is at a maximum or a minimum and indicate which (maximum or minimum, relative or absolute).

I know that two of those points are $x = -2$ and $x = 2$ but my book states that $x=-3$ is also a relative minimum.

However, I don't understand why. My book states that:

In an open (non-empty) interval the relative extremes can happen:

  • at the zeros of the derivative function where there is a sign change
  • at points of the interval where there is no derivative and the lateral derivatives either have different signs of one of them is zero

Following thay logic, shouldn't $x=3$ be a relative maximum? The sign of the derivative is negative on the left and positive on the right.

And why is $x=-3$ considered a relative minimum? It's derivative is unknown, it's left side derivative is non-existant and the only thing I know about it is that the right side derivative is negative.

To sum up, my questions are:

  1. Shouldn't $x=3$ be a relative maximum? And why not?
  2. Why is $x=-3$ considered a relative minimum?

Your approach relies too much on derivatives, and too little on the definitions. A point $x_0$ is a relative minimum if there is a nbd. $I$ such that $f(x_0) \le f(x)$ for all $x \in I \cap D$ (where $D$ is the domain of the function). Intuitively, $f(-3)$ is the highest value among all nearby.

Looking at the graph, $x=-3$ is a relative maximum because $f(-3) \ge f(x)$ for all $x$ in a nbd. of $-3$ and in the domain of $f$. Instead, $x=3$ is not a relative extremum (neither maximum nor minimum) because $f(x^-) < f(3) < f(x^+)$ where $x^-$ is a point in a (sufficiently small) nbd. of $3$ and $x^+$ is a point in a (sufficiently small) nbd. of $3$. Intuitively, $f(3)$ is smaller than the function values on its right and larger than the values on its left.


$x_0=3$ is neither relative minimum, nor relative maximum. If there were $f(3)=0$, then it was a relative minimum. In the situation as at the graph, for every neigobourhood of $3$ we have both $f(x)<f(3)$ and $f(y)>f(3)$ for some $x,y$ close enough to $3$.


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