# Math Insight

### Local minima and maxima (First Derivative Test)

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

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