Increasing/Decreasing | Teaching Calculus

In my last post we discussed the five shapes of a graph. Hopefully, that activity, which is posted under the Resources tab above, helped your students discover that

A function is increasing and concave up, on any interval where its first derivative is positive and its second derivative is positive, like y = sin(x) on $\left( \tfrac{3\pi }{2},2\pi \right)$ .
A function is increasing and concave down, on any interval were its first derivative is positive and its second derivative is negative, like y = sin(x) on $\left( 0,\tfrac{\pi }{2} \right)$ .
A function is decreasing and concave up, on any interval where its first derivative is negative and its second derivative is positive, like y = sin(x) on $\left( \pi ,\tfrac{3\pi }{2} \right)$ .
A function is decreasing and concave down, on any interval where its first derivative is negative and its second derivative is negative, like y = sin(x) on $\left( \tfrac{\pi }{2},\pi \right)$ .
To which we will add a function is linear where its first derivative is constant and its second derivative is zero.

Separating the increasing/decreasing behavior from the concavity:

On an interval where the first derivative is positive the graph is increasing, and on an interval where the first derivative is negative the function is decreasing.
On an interval where the second derivative is positive the function is concave, on an interval where the second derivative is negative the graph is concave down.

Be careful when presenting the ideas above.

None of them consider what happens if one of the other of the derivatives is zero or undefined. There is an important theorem which says, and we must be careful here, “If for all x in an interval, ${f}'\left( x \right)>0$ , then the function is increasing on that interval.”

True enough, but what about y = x³ on an interval containing the origin? Well, the theorem does not apply, since the derivative is not positive everywhere on the interval. The theorem says nothing about what happens when the derivative is zero, only what happens when it is positive. In such cases we need to return to the definition of increasing (which incidentally does not mention derivatives), to determine that y = x³ is increasing on any interval containing the origin (any interval, anywhere, in fact).

Another thing to be careful of is this: Functions increase or decrease on intervals, not at points. If you are asked if a function is increasing or decreasing at a point, or “Is the velocity increasing when t = ….” interpret the question as asking, “Is there a small open interval containing the point, on which the function is increasing or decreasing.”

Using the derivative to give this kind of information about the graph is a big part of the calculus and one of the important uses of derivatives. We can determine information by working with the equation of the derivative. We can also work from the graph of the derivative. This is often easier since it is easy to tell from the graph when the derivative is positive or negative. From the graph of the derivative, we can also see where the derivative is increasing or decreasing, and this tells us the sign of the second derivative and hence about the concavity of the function. I will discuss this in a later post.

Next: Joining the Pieces

The universal quantifier $\forall$ – for any – for every – for all

Many theorems and definitions in mathematics use the phrases “for any”, “for every” or “for all.” The upside-down A is the symbol. The three phrases all mean the same thing!

For example, we have the definition “A function is increasing on an interval if, and only if, for all pairs of numbers x₁ and x₂ in the interval, if x₁< x₂ then f(x₁) < f(x₂).” Whenever you have a theorem or definition with one, restating it with the other two will help students understand it better: “for all pairs of numbers,” “for any pair of numbers” and “for every pair of numbers.”

Increasing and Decreasing Functions

The symbols in the definition above tell the whole story – sure they do. As with any theorem or definition, use the Rule of Four. The definition above is the analytic part. The graphical part is the obvious – the graph goes up to the right. The numerical part is that as the x-values increase in a table, so do the y-values. The verbal part is the two preceding sentences and all the talking you’re going to have to do to explain this.

The function $y=\sin \left( x \right)$ increases on the closed interval $\left[ -\tfrac{\pi }{2},\tfrac{\pi }{2} \right]$ and the function decreases on the closed interval $\left[ \tfrac{\pi }{2},\tfrac{3\pi }{2} \right]$ . The fact that $\tfrac{\pi }{2}$ is in both intervals is not a problem since it is in the intervals, not at the point, that the function increases or decreases. This is because $\sin \left( \tfrac{\pi }{2} \right)$ is larger than all (every, any) values in $\left[ -\tfrac{\pi }{2},\tfrac{\pi }{2} \right]$ , and also larger than all (any, every) of the values in $\left[ \tfrac{\pi }{2},\tfrac{3\pi }{2} \right]$ .

“Playing” with theorems: You will soon have a theorem that says, “If the derivative of a function is positive on an interval, then the function is increasing on the interval.” Nothing in the paragraph above contradicts this, because the hypothesis says nothing about what is true if the derivative is zero. For this you have to go back to the definition. The converse of this theorem is false. Counterexample: $f\left( x \right)={{x}^{3}}$ is increasing on any (all, every) interval containing the origin, yet $f'\left( 0 \right)=0$ . The AP exams do not make a big deal of this; they accept either open or closed intervals for increasing or decreasing.