A function $f$ has a **local maximum** or **relative maximum** at
a point $x_o$ if the values $f(x)$ of $f$ for $x$ ‘near’ $x_o$ are all
less than $f(x_o)$. Thus, the graph of $f$ near $x_o$ has a *peak*
at $x_o$.
A function $f$ has a **local minimum** or **relative minimum** at
a point $x_o$ if the values $f(x)$ of $f$ for $x$ ‘near’ $x_o$ are all
greater than $f(x_o)$. Thus, the graph of $f$ near $x_o$ has a *trough* at $x_o$.
(To make the distinction clear, sometimes the ‘plain’ maximum and
minimum are called **absolute** maximum and minimum.)

Yes, in both these ‘definitions’ we are tolerating ambiguity about
what ‘near’ would mean, although the peak/trough requirement on the
graph could be translated into a less ambiguous definition. But in any
case we'll be able to execute the procedure given below to *find*
local maxima and minima without worrying over a formal definition.

*This procedure is just a variant of things we've already done to
analyze the intervals of increase and decrease of a function, or to
find absolute maxima and minima.* This procedure starts out the same
way as does the analysis of intervals of increase/decrease, and also
the procedure for finding (‘absolute’) maxima and minima of functions.

To find the local maxima and minima of a function $f$ on an interval $[a,b]$:

- Solve $f'(x)=0$ to find
*critical points*of $f$. - Drop from the list any critical points that aren't in the interval $[a,b]$.
- Add to the list the endpoints (and any points of discontinuity
or non-differentiability): we have an
*ordered*list of special points in the interval: $$a=x_o < x_1 < \ldots < x_n=b$$ - Between each pair $x_i
< x_{i+1}$ of points in the list, choose an
auxiliary point $t_{i+1}$. Evaluate the
*derivative*$f'$ at all the auxiliary points. - For each critical point $x_i$, we have the auxiliary points to
each side of it: $t_i < x_i < t_{i+1}$. There are four cases
*best remembered by drawing a picture!*: - if $f'(t_i)>0$ and $f'(t_{i+1}) < 0$ (so $f$ is
*increasing*to the left of $x_i$ and*decreasing*to the right of $x_i$, then $f$ has a*local maximum*at $x_o$. - if $f'(t_i) < 0$ and $f'(t_{i+1})>0$ (so $f$ is
*decreasing*to the left of $x_i$ and*increasing*to the right of $x_i$, then $f$ has a*local minimum*at $x_o$. - if $f'(t_i) < 0$ and $f'(t_{i+1}) <0 $ (so $f$ is
*decreasing*to the left of $x_i$ and*also decreasing*to the right of $x_i$, then $f$ has*neither*a local maximum nor a local minimum at $x_o$. - if $f'(t_i)>0$ and $f'(t_{i+1})>0$ (so $f$ is
*increasing*to the left of $x_i$ and*also increasing*to the right of $x_i$, then $f$ has*neither*a local maximum nor a local minimum at $x_o$.

The endpoints require separate treatment: There is the auxiliary
point $t_o$ just to the *right* of the left endpoint $a$, and the
auxiliary point $t_n$ just to the *left* of the right endpoint $b$:

- At the
*left*endpoint $a$, if $f'(t_o) < 0$ (so $f'$ is*decreasing*to the right of $a$) then $a$ is a*local maximum*. - At the
*left*endpoint $a$, if $f'(t_o)>0$ (so $f'$ is*increasing*to the right of $a$) then $a$ is a*local minimum*. - At the
*right*endpoint $b$, if $f'(t_n) < 0$ (so $f'$ is*decreasing*as $b$ is approached from the left) then $b$ is a*local minimum*. - At the
*right*endpoint $b$, if $f'(t_n)> 0$ (so $f'$ is*increasing*as $b$ is approached from the left) then $b$ is a*local maximum*.

The possibly bewildering list of possibilities really
shouldn't be bewildering after you get used to them. We are already
acquainted with evaluation of $f'$ at auxiliary points between
critical points in order to see whether the function is increasing or
decreasing, and now we're just applying that information to see
whether the graph *peaks, troughs, or does neither* around each
critical point and endpoints. That is, *the geometric meaning of the
derivative's being positive or negative is easily translated into
conclusions about local maxima or minima.*

Find all the local (=relative) minima and maxima of the
function $f(x)=2x^3-9x^2+1$ on the interval $[-2,2]$: To find critical
points, solve $f'(x)=0$: this is $6x^2-18x=0$ or $x(x-3)=0$, so there
are two critical points, $0$ and $3$. Since $3$ is not in the interval
we care about, we drop it from our list. Adding the endpoints to the
list, we have $$-2 < 0< 2$$ as our ordered list of special points. Let's
use auxiliary points $-1,1$. At $-1$ the derivative is $f'(-1)=24>0$,
so the function is increasing there. At $+1$ the derivative is
$f'(1)=-12 < 0$, so the function is decreasing. Thus, since it is
increasing to the left and decreasing to the right of $0$, it must be
that $0$ is a *local maximum*. Since $f$ is increasing to the
right of the left endpoint $-2$, that left endpoint must give a *local minimum*. Since it is decreasing to the left of the right
endpoint $+2$, the right endpoint must be a *local minimum*.

Notice that although the processes of finding *absolute*
maxima and minima and *local* maxima and minima have a lot in
common, they have essential differences. In particular, the only
relations between them are that *critical points* and *endpoints* (and points of discontinuity, etc.) play a big role in
both, and that the *absolute* maximum is certainly a *local*
maximum, and likewise the *absolute* minimum is certainly a *local* minimum.

For example, just plugging critical points into the function does not
reliably indicate which points are *local* maxima and minima. And,
on the other hand, knowing which of the critical points are *local* maxima and minima generally is only a small step toward
figuring out which are *absolute*: values still have to be plugged
into the function! *So don't confuse the two procedures!*

(By the way: while it's fairly easy to make up story-problems where the
issue is to find the maximum or minimum value of some function on some
interval, it's harder to think of a simple application of *local* maxima
or minima).

#### Exercises

- Find all the local (=relative) minima and maxima of the function $f(x)=(x+1)^3-3(x+1)$ on the interval $[-2,1]$.
- Find the local (=relative) minima and maxima on the interval $[-3,2]$ of the function $f(x)=(x+1)^3-3(x+1)$.
- Find the local (relative) minima and maxima of the function $f(x)=1-12x+x^3$ on the interval $[-3,3]$.
- Find the local (relative) minima and maxima of the function $f(x)=3x^4 - 8x^3 + 6x^2 + 17$ on the interval $[-3,3]$.