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

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.
