Open or Closed?

About this time of year, you find someone, hopefully one of your students, asking, “If I’m finding where a function is increasing, is the interval open or closed?”

Do you have an answer?

This is a good time to teach some things about definitions and theorems.

The place to start is to ask what it means for a function to be increasing. Here is the definition:

A function is increasing on an interval if, and only if, for all (any, every) pairs of numbers x1 < x2 in the interval, f(x1) < f(x2).”

(For decreasing on an interval, the second inequality changes to f(x1) > f(x2). All of what follows applies to decreasing with obvious changes in the wording.)

  1. Notice that functions increase or decrease on intervals, not at individual points. We will come back to this in a minute.
  2. Numerically, this means that for every possible pair of points, the one with the larger x-value always produces a larger function value.
  3. Graphically, this means that as you move to the right along the graph, the graph is going up.
  4. Analytically, this means that we can prove the inequality in the definition.

For an example of this last point consider the function f(x) = x2. Let x2 = x1 + h where h > 0. Then in order for f(x1) < f(x2) it must be true that

\displaystyle {{x}_{1}}^{2}<{{\left( {{{x}_{1}}+h} \right)}^{2}}

\displaystyle 0<{{\left( {{{x}_{1}}+h} \right)}^{2}}-{{x}_{1}}^{2}

\displaystyle 0<{{x}_{1}}^{2}+2{{x}_{1}}h+{{h}^{2}}-{{x}_{1}}^{2}

\displaystyle 0<2{{x}_{1}}h+{{h}^{2}}

This can only be true if \displaystyle {{x}_{1}}\ge 0 Thus, x2 is increasing only if\displaystyle {{x}_{1}}\ge 0

Now, of course, we rarely, if ever, go to all that trouble. And it is even more trouble for a function that increases on several intervals.  The usual way of finding where a function is increasing is to look at its derivative.

Notice that the expression \displaystyle 0<{{x}_{1}}^{2}+2{{x}_{1}}h+{{h}^{2}}-{{x}_{1}}^{2} looks a lot like the numerator of the original limit definition of the derivative of x2 at x = x1, namely \displaystyle 0<{{\left( {{{x}_{1}}+h} \right)}^{2}}-{{x}_{1}}^{2}. If h > 0, where the function is increasing the numerator is positive and the derivative is positive also. Turning this around we have a theorem that says, If \displaystyle {f}'\left( {{{x}_{1}}} \right)>0 for all x in an interval, then the function is increasing on the interval. That makes it much easier to find where a function is increasing, we simplify find where its derivative is positive.

There is only a slight problem in that the theorem does not say what happens if the derivative is zero somewhere on the interval. If that is the case, we must go back to the definition of increasing on an interval or use a different method. For example, the function x3 is increasing everywhere, even though its derivative at the origin is zero.

Let’s consider another example. The function sin(x) is increasing on the interval \displaystyle [-\tfrac{\pi }{2},\tfrac{\pi }{2}] (among others) and decreasing on \displaystyle [\tfrac{\pi }{2},\tfrac{3\pi }{2}]. It bothers some that \displaystyle \tfrac{\pi }{2} is in both intervals and that the derivative of the function is zero at x = \displaystyle \tfrac{\pi }{2}. This is not a problem. Sin(\displaystyle \tfrac{\pi }{2}) is larger than all the other values is both intervals, so by the definition, and not the theorem, the intervals are correct.

It is generally true that if a function is continuous on the closed interval [a,b] and increasing on the open interval (a,b) then it must be increasing on the closed interval [a,b] as well.

Returning to the first point above: functions increase or decrease on intervals not at points. You do find questions in books and on tests that ask, “Is the function increasing at x = a.” The best answer is to humor them and answer depending on the value of the derivative at that point. Since the derivative is a limit as h approaches zero, the function must be defined on some interval around x = a in which h is approaching zero. So, answer according to the value of the derivative on that interval.

You can find more on this here.

Case Closed.

Slightly revised from a posted published on November 2, 2012.

Getting Ready

Today we continue to look at some previous posts that I hope will help you and your students throughout the year. We begin with some posts on graphing calculator use and then a few general things in three posts on beginning the year, followed by some mathematics I hope students know before they start studying the calculus.

Graphing Calculators

There are four things that students may, and are required to know how to do, for the AP Exams. But graphing calculators are not required just to answer a few questions on the exams. They are to encourage investigations and experimentation in all math classes. And not just graphing calculator use but all kinds of appropriate technology. So, don’t restrict yourself and your students to only those operations required on the exam. That said, here are previous posts on exam calculator use; as the year goes on there will be other posts on the use of graphing calculators and other technology in your class.

Starting School

A little late perhaps …

These posts discuss basic ideas that I always hoped students knew about mathematics before starting calculus

 

 

 

.

Definitions 2

In helping students read and understand mathematics knowing about definitions, axioms (aka assumptions, postulates) and theorems. By this I mean knowing the parts of a definition or theorem and how they relate to each other should increase the students’ understanding. Today I’ll discuss definitions; theorems and assumptions will be discussed in a future post.

A definition names some mathematical “thing.” A good definition (in mathematics or elsewhere) names the thing defined in a sentence. The sentence may contain symbols, which are really just shorthand for words. A definition has 4 characteristics:

  1. It should put the thing defined into the nearest group of similar things.
  2. It should give the characteristics that distinguish it from the other things in the group.
  3. It should use simpler terms (previously defined terms).
  4. It should be reversible.

I will discuss each of these with an example first from geometry and then from calculus. First however, a word or two about “reversible.” Definitions are what are known technically as bi-conditional statement, meaning that the statement and its converse are true. More on this in the next post.

An example from geometry:

Definition: An equilateral triangle is a triangle with three congruent sides.

The term defined is “equilateral triangle.”

  1. Nearest group of similar things: triangles
  2. Distinguishing characteristic: 3 congruent sides. We all know that an equilateral triangle also has 3 congruent angles, and that all the angles have a measure of \displaystyle \tfrac{\pi }{3}, and all the angles add up to a straight angle, and lots of other great things, but for the definition we only mention the feature that distinguishes equilateral triangles from other triangles. It would be possible to use instead the 3 congruent angles or the fact that all three angles measure are \displaystyle \tfrac{\pi }{3}, as the distinguishing characteristic, but whoever wrote the definition choose the sides. (We could not use the fact that the angles add to a straight angle, because that is true for all triangles and therefore doesn’t distinguish equilateral triangles from the others.) Definitions do not list all the things that may be true, only those that make it different.
  3. Simpler terms: triangle, sides (of a triangle) and congruent. We assume that these key terms are already known to the student. Of course there were no previously defined terms for the very first things (points, lines and planes) but by now we are past that and have lots of previously defined terms to work with.
  4. Reversible: If we know that this object is an equilateral triangle, then without looking further we know it has 3 congruent sides AND if we run across a triangle with 3 congruent sides, we know it must be an equilateral triangle.

An example from the calculus:

Definition: A function, f, is increasing on an interval if, and only if, for all pairs of numbers a and b in the interval, if a < b, then f (a) < f (b).

This is a little more complicated. The term being defined is increasing on an interval. This becomes important and can lead to confusion because sometimes we are tempted to think functions are increasing at a point. There is no definition for the latter: functions increase only on intervals.

  1. Nearest group of similar things: functions
  2. Distinguishing characteristic: for all pairs of numbers a and b in the interval, if a < b, then f (a) < f (b).
    1. The if …, then … construction indicates a conditional statement (discussed in the next post) inside of the definition. This is not uncommon. It means that if can establish that this is true, then we can say then function is increasing on the interval.
    2. The phrase “for all” is also common in mathematics. It means the same thing as “for any” and “for every.” When you come across one of these it is a very good idea to rephrase the sentence with each of them: “for all numbers a and b in the interval…”, “for any pair of numbers a and b in the interval …” and “for every two numbers a and b in the interval…” This greatly helps understanding definitions.
    3. Simpler terms: function, interval (could be open, closed or half-open), less than (<), the meaning of symbols like f (a).
    4. Reversible:
      1. the phrase “if, and only if” indicates that what goes before and what comes after it, each imply the other. This phrase is implicit in all (any, every) definitions although English usage often omits it. The first definition could be written “A triangle is equilateral if, and only if, it has 3 congruent sides” but is a little more user-friendly the way it is stated above.
      2. If you can establish that “for all pairs of numbers a and b in the interval, if a < b, then f (a) < f (b)”, then you can be sure the function increases on the interval. AND if you are told f is increasing on the interval, then without checking further you can be sure that “for all (any, every) pairs of numbers a and b in the interval, if a < b, then f (a) < f (b).”

Now that’s a fairly detailed discussion (definition?) of a definition. But it is worth going through any new definition for your students to help them learn what the definition really means. First identify the four features for you students and then as new definitions come along have them identify the parts. Encourage them to pull definitions apart this way. It is worth the little extra time spent.

Open or Closed?

About this time of year you find someone, hopefully one of your students, asking, “If I’m finding where a function is increasing, is the interval open or closed?”

Do you have an answer?

This is a good time to teach some things about definitions and theorems.

The place to start is to ask what it means for a function to be increasing. Here is the definition:

A function is increasing on an interval if, and only if, for all (any, every) pairs of numbers x1 < x2 in the interval, f(x1) < f(x2).

(For decreasing on an interval, the second inequality changes to f(x1) > f(x2). All of what follows applies to decreasing with obvious changes in the wording.)

  1. Notice that functions increase or decrease on intervals, not at individual points. We will come back to this in a minute.
  2. Numerically, this means that for every possible pair of points, the one with the larger x-value always produces a larger function value.
  3. Graphically, this means that as you move to the right along the graph, the graph is going up.
  4. Analytically, this means that we can prove the inequality in the definition.

For an example of this last point consider the function f(x) = x2. Let x2 = x1 + h where h > 0. Then in order for  f(x1) < f(x2) it must be true that

{{x}_{1}}^{2}<{{\left( {{x}_{1}}+h \right)}^{2}}
0<{{\left( {{x}_{1}}+h \right)}^{2}}-{{x}_{1}}^{2}
0<{{x}_{1}}^{2}+2h{{x}_{1}}+{{h}^{2}}-{{x}_{1}}^{2}
0<h\left( 2{{x}_{1}}+h \right)

This can only be true if {{x}_{1}}\ge 0, Thus, x2 is increasing only if x\ge 0.

Now, of course, we rarely, if ever, go to all that trouble. And it is even more trouble for a function that increases on several intervals.  The usual way of finding where a function is increasing is to look at its derivative.

Notice that the expression {{\left( {{x}_{1}}+h \right)}^{2}}-{{x}_{1}}^{2} looks a lot like the numerator of the original limit definition of the derivative of x2 at x = x1, namely \displaystyle {f}'\left( {{x}_{1}} \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( {{x}_{1}}+h \right)}^{2}}-{{x}_{1}}^{2}}{h}. If h > 0, where the function is increasing the numerator is positive and the derivative is positive also. Turning this around we have a theorem that says, If {f}'\left( x \right)>0 for all x in an interval, then the function is increasing on the interval. That makes it much easier to find where a function is increasing: we simplify find where its derivative is positive.

There is only a slight problem in that the theorem does not say what happens if the derivative is zero somewhere on the interval. If that is the case, we must go back to the definition of increasing on an interval or use some other method. For example, the function x3 is increasing everywhere, even though its derivative at the origin is zero.

Let’s consider another example. The function sin(x) is increasing on the interval \left[ -\tfrac{\pi }{2},\tfrac{\pi }{2} \right] (among others) and decreasing on \left[ \tfrac{\pi }{2},\tfrac{3\pi }{2} \right]. It bothers some that \tfrac{\pi }{2} is in both intervals and that the derivative of the function is zero at x = \tfrac{\pi }{2}. This is not a problem. Sin(\tfrac{\pi }{2}) is larger than all the other values is both intervals, so by the definition, and not the theorem, the intervals are correct.

It is generally true that if a function is continuous on the closed interval [a,b] and increasing on the open interval (a,b) then it must be increasing on the closed interval [a,b] as well. (There is a proof by Lou Talman of this fact click here .)

Returning to the first point above: functions increase or decrease on intervals not at points. You do find questions in books and on tests that ask, “Is the function increasing at x = a.” The best answer is to humor them and answer depending on the value of the derivative at that point. Since the derivative is a limit as h approaches zero, the function must be defined on some interval around x = a in which h is approaching zero. So answer according to the value of the derivative on that interval.

You can find more on this here.

Case Closed.

Definitions

Definitions are similar to theorems, but are true in both directions; technically, this means that the statement and its converse are both true (p\leftrightarrow q). The double arrow is read “if, and only if.” Both parts are either true or both parts are false. Definitions usually name some thing or some property.  Definitions are not proved.

The definition of continuity is a good example: A function f is continuous at xa if, and only if, these three things are true

(1)  f\left( a \right) exist (i.e. is a finite number)

(2)  \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) exist (i.e. is a finite number)

(3) \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)  (“The limit equals the value.”)

“Play” with it: consider cases where only 2 of the 3 requirements are true – is the function still continuous? What would happen if you removed the requirements about finite numbers?

To use a theorem, one must be sure all the hypotheses are true. To use a definition, one may say that either part is true once you have established that the other part is true. So, if you know a function is continuous at a point, then the three statements are true; or if you can show the three statements are true, you may say the function is continuous.

Here’s an example: A typical AP problem might give a piecewise defined function and ask if it is continuous at the place where the domain is divided.

To get credit for justifying an answer of “yes”, students must show that all the requirements of the definition are met. Specifically, they must show that the limit as x approaches that point must equal the value of  the function at that point (and both are finite).  In turn, to show that this limit exist the student must show that the hypotheses of the theorem that says if the two one-sided limits are equal to the same number, then that number is the limit.

To get credit for an answer of “no”, the student must show that (only) one of the hypotheses is false.

Finally, as with theorems, express definitions in words. With your students, “play” with the theorem or definition by making changes to the hypotheses and seeing how that affects the conclusion. Look at the graphs. Don’t just state the definition and expect students to understand it, remember it and use it correctly.