# Math Insight

### 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

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