My textbook has these passages written and I'm a bit confused as to whether local maximums and minimums can be on endpoints. why or why not?
Apparently x=4 is not a local maximum. But why?
Other passages:
[Math] can local maximums and minimums be on endpoints
calculus
Related Solutions
I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.
edit 1:(add more details)
$f(x,y)=(x-2)^2+y^2+5 \implies (x-2)^2+y^2=f(x,y)-5=R^2$
you want to find max and min of $f(x,y)$ is same to find max and min of $R$.
note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$
if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.
Actually, the question is settled by reading the definition you provided carefully:
A function $f$ has local maximum value at point $c$ within its domain $D$ if $f(x)\leq f(c)$ for all $x$ in its domain lying in some open interval containing $c$.
I.e., the points $x$ for which the condition must hold are required to both be in the open interval and in $D$.
To see that $(1,1)$ is a local maximum, consider the open interval $(0, 2)$. If $x \in (0, 2)$ and $x$ is also in the domain $[1,\infty)$, then $1 \le x < 2$. Now $g(x) = x^2-4x + 4 = (2 - x)^2$. So $g(1) = (2 - 1)^2 = 1^2 = 1$, but if $x > 1$, then $0 < 2 - x < 1$, so $0 < (2-x)^2 = g(x) < 1$. So for $x$ in the open interval $(0,2)$ and also in the domain $[1,\infty)$, we have that $g(x) \le g(1)$.
Best Answer
Not according to the definition in your second scanned image.
Short answer: because the "near $c$" language in the definition means "near $c$" on both sides of $c$.
Longer answer: According to the definition in the red box of your second scanned image,
where "near $c$" means
However, for the function $f$ whose graph is plotted in Figure 7, every open interval containing $4$ will contain some values $x > 4$, for which $f(x)$ is not defined. (The domain of $f$ was given to be the closed interval $[-1, 4]$.) So, it's not possible to have $f(4) \geq f(x)$ for every $x$ in some open interval containing $4$.
The discussion above addresses the definition of local maximum, but the definition for local minimum is analogous and has the same issue.