2
$\begingroup$

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?

enter image description here

Apparently x=4 is not a local maximum. But why?

Other passages:

enter image description here

$\endgroup$
2
$\begingroup$

can local maximums and minimums be on endpoints?

Not according to the definition in your second scanned image.

Apparently $x=4$ is not a local maximum. But why?

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,

the number $f(c)$ is a local maximum value of $f$ if $f(c) \geq f(x)$ when $x$ is near $c$

where "near $c$" means

on some open interval containing $c$.

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.

$\endgroup$
0
$\begingroup$

When you look for optima of a function on a bounded interval (as opposed to the entire space), the optimum point can come from

  1. Local extrema inside the boundary
  2. Boundary itself.

A good example is $f(x) = x^2$ on $I=[-1,2]$ which has a unique minimum on $\mathbb{R}$, which happens to fall inside $I$, but for the maximum, the situation is more tricky. $f$ has no local maximum over $\mathbb{R}$ but the boundary is limited to $2$ points $(-1,1)$ and $(2,4)$, both of which are candidates for the optimum and indeed it occurs at $(2,4)$.

$\endgroup$
  • 1
    $\begingroup$ So what is the answer to OP's question? $\endgroup$ – ziggurism Feb 20 '18 at 4:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.