### Critical points, monotone increase and decrease

A function is called **increasing** if it increases as the input $x$
moves from left to right, and is called **decreasing** if it
decreases as $x$ moves from left to right. Of course, a function can
be increasing in some places and decreasing in others: that's the
complication.

We can notice that a function is increasing if the slope of its
tangent is positive, and decreasing if the slope of its tangent is
negative. Continuing with the idea that the slope of the tangent is the
derivative: *a function is increasing where its derivative is
positive, and is decreasing where its derivative is negative*.

This is a great principle, because we don't have to graph the function or otherwise list lots of values to figure out where it's increasing and decreasing. If anything, it should be a big help in graphing to know in advance where the graph goes up and where it goes down.

And the points where the tangent line is horizontal, that is, where
the derivative is zero, are **critical points**. The points where
the graph has a *peak* or a *trough* will certainly lie among
the critical points, although there are other possibilities for
critical points, as well.

Further, for the kind of functions we'll deal with here, there is a fairly systematic way to get all this information: to find the intervals of increase and decrease of a function $f$:

- Compute the derivative $f'$ of $f$, and
*solve*the equation $f'(x)=0$ for $x$ to find all the critical points, which we list in order as $x_1 < x_2 < \ldots < x_n$. - (If there are points of discontinuity
or non-differentiability, these points should be added to the list!
But points of discontinuity or non-differentiability are
*not*called*critical points*.) - We need some
*auxiliary points*: To the left of the leftmost critical point $x_1$ pick any convenient point $t_o$, between each pair of consecutive critical points $x_i,x_{i+1}$ choose any convenient point $t_i$, and to the right of the rightmost critical point $x_n$ choose a convenient point $t_n$. - Evaluate the
*derivative*$f'$ at all the*auxiliary*points $t_i$. - Conclusion: if $f'(t_{i+1})>0$, then $f$ is
*increasing*on $(x_i,x_{i+1})$, while if $f'(t_{i+1}) <0$, then $f$ is*decreasing*on that interval. - Conclusion: on the ‘outside’ interval $(-\infty,x_o)$, the
function $f$ is
*increasing*if $f'(t_o)>0$ and is*decreasing*if $f'(t_o) <0$. Similarly, on $(x_n,\infty)$, the function $f$ is*increasing*if $f'(t_n)>0$ and is*decreasing*if $f'(t_n) <0$.

It is certainly true that there are many possible shortcuts to this procedure, especially for polynomials of low degree or other rather special functions. However, if you are able to quickly compute values of (derivatives of!) functions on your calculator, you may as well use this procedure as any other.

Exactly which *auxiliary points* we choose does not matter, as
long as they fall in the correct intervals, since we just need a
single sample on each interval to find out whether $f'$ is positive or
negative there. Usually we pick integers or some other kind of number
to make computation of the derivative there as easy as possible.

It's important to realize that even if a question does not directly
ask for *critical points*, and maybe does not ask about *intervals* either, still it is *implicit* that we have to find the
critical points and see whether the functions is increasing or
decreasing on the *intervals between critical points*.

#### Examples

Find the critical points and intervals on which
$f(x)=x^2+2x+9$ is increasing and decreasing: Compute
$f'(x)=2x+2$. Solve $2x+2=0$ to find only one critical point $-1$. To
the left of $-1$ let's use the *auxiliary point* $t_o=-2$ and to
the right use $t_1=0$. Then $f'(-2)=-2
<0$, so $f$ is *decreasing*
on the interval $(-\infty,-1)$. And $f'(0)=2>0$, so $f$ is *increasing* on the interval $(-1,\infty)$.

Find the critical points and intervals on which
$f(x)=x^3-12x+3$ is increasing, decreasing. Compute
$f'(x)=3x^2-12$. Solve $3x^2-12=0$: this simplifies to $x^2-4=0$, so
the *critical points* are $\pm 2$. To the left of $-2$ choose *auxiliary point* $t_o=-3$, between $-2$ and $+2$ choose auxiliary
point $t_1=0$, and to the right of $+2$ choose $t_2=3$. Plugging in
the auxiliary points to the derivative, we find that $f'(-3)=27-12>0$,
so $f$ is *increasing* on $(-\infty,-2)$. Since $f'(0)=-12
<0$, $f$
is *decreasing* on $(-2,+2)$, and since $f'(3)=27-12>0$, $f$ is
*increasing* on $(2,\infty)$.

Notice too that we don't really need to know the exact value of the derivative at the auxiliary points: all we care about is whether the derivative is positive or negative. The point is that sometimes some tedious computation can be avoided by stopping as soon as it becomes clear whether the derivative is positive or negative.

#### Exercises

- Find the critical points and intervals on which $f(x)=x^2+2x+9$ is increasing, decreasing.
- Find the critical points and intervals on which $f(x)=3x^2-6x+7$ is increasing, decreasing.
- Find the critical points and intervals on which $f(x)=x^3-12x+3$ is increasing, decreasing.

#### Thread navigation

##### Calculus Refresher

- Previous: Tangent and normal lines
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