Finding the local extreme values of $f(x) = -x^2 + 2x + 9$ over $[-2,\infty)$.

Question

I'm tutoring a student, and we were trying to solve the following question:

Find the local extreme values of $f(x) = -x^2 + 2x + 9$ over $[-2,\infty)$.

According to the textbook, the local extreme values are essentially the peaks and the valleys in the graph of the function $f$, so basically where $f'(x) = 0$. This is relatively easy to compute: $$f'(x) = -2x + 2,$$ of which the critical points are $x = 1$. Likewise, $f''(x) = -2 < 0$, which means $f$ is concave down everywhere, and thus $x = 1$ is where a maximum value occurs on the graph. The maximum is $f(1) = -1 + 2 + 9 = 10$.

Of course, the endpoint $x = -2$ yields $$f(-2) = -(-2)^2 +2(-2) + 9 = 1,$$ but since the graph is concave down everywhere, $\displaystyle \lim_{x\to\infty}f = -\infty$ implies there really is no minimum per se... right?

The online computer program tells us that $(-2,1)$ is a local minimum, and $(1,10)$ is a local maximum. But in accordance with the definition from the textbook, why is $(-2,1)$ where a local minimum of the graph occurs? It's neither a peak nor a valley in the graph. What exactly does local mean when the interval is infinite? It doesn't quite make logical sense, unless the definition is not as rigorous as it ought to be.

Answer 1

Sometimes just plotting the function cuts through needless distractions:

Clearly at the point $x=-2$, the function is lower than any other values in the specified domain, and hence is a local minimum.

Answer 2

If you look at the statements of the first and second derivative tests, they should indicate that they are valid at interior points of intervals, not the endpoints. (And only where the function happens to be differentiable.) The endpoints are a special case. An endpoint can be a max or min while the derivatives there can be anything at all. A local max or min is defined to be the biggest or smallest guy in his neighborhood. Period.

Second, the key word is "local". The function here doesn't have an "absolute" (or "global") min, but it has a local min.

Answer 3

This is a parabola open downward.

The vertex is $(1,10)$ where your derivative is zero and it is a local and global maximum.

The left endpoint of $(-2, 1)$ is a local minimum and there is no global minimum.

Answer 4

According to this Wikipedia page, the proper definition of local maxima and local minima is:

$f:X\longmapsto$Y is said to have a local (or relative) maximum point at the point $x^∗$ if there exists some $ε > 0$ such that $f(x^∗) ≥ f(x)$ for all $x\in$ X within distance $ε$ of $x^∗$. Similarly, the function has a local minimum point at $x^∗$ if $f(x^∗) ≤ f(x)$ for all $x\in$ X within distance $ε$ of $x^∗$.

Since we can fix a $\epsilon$-neighborhood with our choice of $\epsilon$ near $x=-2$ so that in this domain $(-2, 1)$ is a minimum point, it is indeed a local minimum.

Answer 5

For the purposes of local extrema we do not consider the endpoints of the interval in question. Thus the only local extremum (in this case a local maxiumum) occurs at $(1, 10)$. On the interval in question, $(1, 10)$ also happens to be an absolute maximum, and there are no local nor absolute minima on this interval.