# Differentiability Implies Continuity

An important theorem concerning derivatives is this:

If a function f is differentiable at x = a, then f is continuous at x = a.

The proof begins with the identity that for all $x\ne a$

$\displaystyle f\left( x \right)-f\left( a \right)=\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}$

$\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {x-a} \right)\cdot \underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}$

$\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=0\cdot {f}'\left( a \right)=0$

And therefore, $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right)$

Since both sides are finite, the function is continuous at x = a.

The converse of this theorem is false: A continuous function is not necessarily differentiable. A counterexample is the absolute value function which is continuous at the origin but not differentiable there. (The slope approaching from the left is not equal to the slope from the right.)

This is a theorem whose contrapositive is used as much as the theorem itself. The contrapositive is,

If a function is not continuous at a point, then it is not differentiable there.

Example 1: A function such as  $\displaystyle g\left( x \right)=\frac{{{{x}^{2}}-9}}{{x-3}}$ has a (removable) discontinuity at x = 3, but no value there.

So, in the limit definition of the derivative, $\displaystyle \text{ }\!\!~\!\!\text{ }\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{g\left( {3+h} \right)-g\left( 3 \right)}}{h}$ there is no value of g(3) to use, and the derivative does not exist.

Example 2:  $\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}} & {x\le 1} \\ {{{x}^{2}}+3} & {x>1} \end{array}} \right.$. This function has a jump discontinuity at x = 1.

Since the point (1, 1) is on the left part of the graph, if h > 0, $f\left( {1+h} \right)-f\left( 1 \right)>3$ and the limit  will always be a number greater than 3 divided by zero and will not exist. Therefore, even though the slopes from both side of x =1 approach the same value, namely 2, the derivative does not exist at x = 1.

This also applies to a situation like example 1 if f(3) were some value that did not fill in the hole in the graph.

On the AP Calculus exams students are often asked about the derivative of a function like those in the examples, and the lack of continuity should be an immediate clue that the derivative does not exist. See 2008 AB 6 (multiple-choice).

Just as important are questions in which the function is given as differentiable, but the student needs to know about continuity. Just remember: differentiability implies continuity. See 2013 AB 14 in which you must realize the since the function is given as differentiable at x = 1, it must be continuous there to solve the problem.

Continuity of the Derivative

A question that comes up is, if a function is differentiable is its derivative differentiable? The answer is no. While almost always the derivative is also differentiable, there is this counterexample:

$\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}\sin \left( {\frac{1}{x}} \right)} & {x\ne 0} \\ 0 & {x=0} \end{array}} \right.$

The first line of the function has a removable oscillating discontinuity at x = 0, but since the $\displaystyle {{x}^{2}}$ factor squeezes the function to the origin; the added condition that $\displaystyle f\left( 0 \right)=0$ makes the function continuous. Differentiating gives

$\displaystyle {{f}^{'}}\left( x \right)={{x}^{2}}\cos \left( {\frac{1}{x}} \right)\left( {\frac{{-1}}{{{{x}^{2}}}}} \right)+2x\sin \left( {\frac{1}{x}} \right)=-\cos \left( {\frac{1}{x}} \right)+2x\sin \left( {\frac{1}{x}} \right)$

And now there is no way to get around the oscillating discontinuity at x = 0.

# Then there is this – Existence Theorems

Existence Theorems

An existence theorem is a theorem that says, if the hypotheses are met, that something, usually a number, must exist.

For example, the Mean Value Theorem is an existence theorem: If a function f is defined on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in the open interval (a, b) such that $\displaystyle {f}'\left( c \right)\left( {b-a} \right)=f\left( b \right)-f\left( a \right)$.

The phrase “there exists” can also mean “there is” and “there is at least one.” In fact, it is a good idea when seeing an existence theorem to reword it using each of these other phrases. “There is at least one” reminds you that there may be more than one number that satisfies the condition. The mathematical symbol for these phrases is an upper-case E written backwards: $\displaystyle \exists$.

Textbooks, after presenting an existence theorem, usually follow-up with some exercises asking students to find the value for a given function on a given interval: “Find the value of c guaranteed by the Mean Value Theorem for the function … on the interval ….” These exercises may help students remember the formula involved but are not very useful otherwise.

The important thing about most existence theorems is that the number exists, not what the number is. To illustrate this, let’s consider the Fundamental Theorem of Calculus. After partitioning the interval [a, b] into subintervals at various values, xi, we consider the limit of the sum

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{n=1}}^{n}{{\left( {f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)} \right)}}$.

Write out a few terms and you will see that is a telescoping series and its limit is $\displaystyle f\left( b \right)-f\left( a \right)$.

The expression $\displaystyle {f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)}$ resembles the right side of the Mean Value Theorem above. Since all the conditions are met, the MVT tells us that in each subinterval $\displaystyle [{{x}_{{i-1}}},{{x}_{i}}]$ there exists a number, call it ci , such that

$\displaystyle {f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-1}}}} \right)=f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)$ and therefore

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{n=1}}^{n}{{\left( {f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)} \right)}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{n=1}}^{n}{{{f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-1}}}} \right)}}=f\left( b \right)-f\left( a \right)$

No one is concerned what these ci are, just that there are such numbers, that they exist. (The second limit above is then defined as the definite integral so $\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{n=1}}^{n}{{{f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-1}}}} \right)}}=\int_{a}^{b}{{{f}'\left( x \right)dx=}}f\left( b \right)-f\left( a \right)$ – The Fundamental Theorem of Calculus.)

Other important existence theorems in calculus

The Intermediate Value Theorem

If f is continuous on the interval [a, b] and M is any number between f(a) and f(b), then there exists a number c in the open interval (a, b) such that f(c) = M.

If f is continuous on an interval and f changes sign in the interval, then there must be at least one number c in the interval such that f(c) = 0

Extreme Value Theorem

If f is continuous on the closed interval [a, b], then there exists a number c in [a, b] such that $\displaystyle f\left( c \right)\ge f\left( x \right)$ for all x in the interval. Every function continuous on a closed interval has (i.e. there exists) a maximum value in the interval.

If f is continuous on the closed interval [a, b], then there exists a number c in [a, b] such that $\displaystyle f\left( c \right)\le f\left( x \right)$ for all x in the interval. Every function continuous on a closed interval has (i.e. there exists) a minimum value in the interval.

Critical Points

If f is differentiable on a closed interval and $\displaystyle {f}'\left( x \right)$ changes sign in the interval, then there exists a critical point in the interval.

Rolle’s theorem

If a function f is defined on the closed interval [a, b] and differentiable on the open interval (a, b) and f(a) = f(b), then there must exist a number c in the open interval (a, b) such that $\displaystyle {f}'\left( c \right)=0$.

MVT – other forms

If I drive a car continuously for 150 miles in three hours, then there is a time when my speed was exactly 50 mph.

If a function f is defined on the closed interval [a, b] and differentiable on the open interval (a, b), then there is a point on the graph of f where the tangent line is parallel to the segment between the endpoints.

Taylor’s Theorem

If f is a function with derivatives through order n + 1 on an interval I containing a, then, for each x in I , there exists a number c between x and a such that

$\displaystyle f\left( x \right)=\sum\limits_{k=0}^{n}{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}+\frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}}$

The number $\displaystyle R=\frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}}$ is called the remainder. The equation above says that if you can find the correct c the function is exactly equal to Tn(x) + R. Notice the form of the remainder is the same as the other terms, except it is evaluated at the mysterious c. The trouble is we almost never can find the c without knowing the exact value of f(x), but; if we knew that, there would be no need to approximate. However, often without knowing the exact values of c, we can still approximate the value of the remainder and thereby, know how close the polynomial Tn(x) approximates the value of f(x) for values in x in the interval, i. See Error Bounds and the Lagrange error bound.

Cogito, ergo sum

And finally, we have Descartes’ famous “theorem” Cogito, ergo sum (in Latin) or the original French, Je pense, donc je suis, translated as “I think, therefore I am” proving his own existence.

# Y the FTC?

So, you’ve finally proven the Fundamental Theorem of Calculus and have written on the board:

$\displaystyle \int_{a}^{b}{{{f}'\left( x \right)dx=f\left( b \right)-f\left( a \right)}}$

And the students ask, “What good is that?” and “When are we ever going to use that?” Here’s your answer.

There are two very important uses of this theorem. Show them BOTH uses right away to help your students see why the FTC is so useful and important.

First, in words the theorem says that “the integral of a rate of change is the net amount of change.” So, if you are given a rate of change (as you are every year on the AP Calculus exam) and asked to find the amount of change (as you are every year on the AP Calculus exam), this is what you use, Show an example such as 2015 AB 1/BC1 that states,

“The rate at which rainwater flows into a drain pipe is modeled by the function R, where $R\left( t \right)=20\sin \left( {\frac{{{{t}^{2}}}}{{35}}} \right)$ cubic feet per hour….

“(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval ?”

The answer is of course, $\displaystyle \int_{0}^{8}{{20\sin \left( {\frac{{{{t}^{2}}}}{{35}}} \right)dt}}$. (Which they will soon learn how to evaluate.)

Second, a more immediate use is to avoid all that work you’ve been doing setting up Riemann sums and finding their limits. No more of that! Give them this integral to evaluate:

$\displaystyle \int_{2}^{7}{{2xdx}}$

Draw the trapezoid representing the area between the graph of $y=2x$ and the x-axis on the interval [2,7] and find its area =  $\displaystyle \frac{1}{2}\left( 5 \right)\left( {18+4} \right)=45$

Then ask, “Does anyone know of a function whose derivative is $2x$?” Let them think for a minute and someone will say, “Yeah, it’s ${{x}^{2}}$”  And then show them

$\displaystyle \int_{2}^{7}{{2xdx}}={{7}^{2}}-{{2}^{2}}=45$

Then go for a harder one:  $\displaystyle \int_{0}^{{\frac{\pi }{2}}}{{\cos \left( x \right)dx}}$

“Does anyone know a function whose derivative is $\cos \left( x \right)$?”

“Why yes, it’s $\sin \left( x \right)$

So, $\displaystyle \int_{0}^{{\frac{\pi }{2}}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{\pi }{2}} \right)-\sin \left( 0 \right)=1-0=1$

That was easy!

If you want to challenge them and review some functions of the “special angles” try this one:

$\displaystyle \int_{{\frac{\pi }{6}}}^{{\frac{{4\pi }}{3}}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{{4\pi }}{3}} \right)-\sin \left( {\frac{\pi }{6}} \right)=\frac{{\sqrt{3}}}{2}-\frac{1}{2}$

Tie the two parts together: Look at the graph of $y=\sin \left( x \right)$. How much does it change from 0 to $\frac{\pi }{2}$? How much does it change from $\frac{\pi }{6}$ to $\frac{{4\pi }}{3}$?

1. “The function whose derivative is …” is called the antiderivative.
2. Using antiderivatives to evaluate definite integrals is easy; the hard part is finding the antiderivatives, since they are not all as straightforward as the two examples above. So, next we need to spend a few weeks learning how to find antiderivatives.[1]
3. Given a derivative, finding its antiderivative is also the start of solving differential equations. This, too, will soon be a concern in the course.

[1] As I’ve written before, this is where it seems logical place to teach antiderivatives. Now students have a reason to find them. Teaching antidifferentiation after differentiation, before integration, seems like an intellectual exercise. Sure, it’s fun, but now we have a need for it.

# Inverses

This series of posts reviews and expands what students know from pre-calculus about inverses. This leads to finding the derivative of exponential functions, ax, and the definition of e, from which comes the definition of the natural logarithm.

Inverses Graphically and Numerically

The Range of the Inverse

The Calculus of Inverses

The Derivatives of Exponential Functions and the Definition of e and This pair of posts shows how to find the derivative of an exponential function, how and why e is chosen to help this differentiation.

Logarithms Inverses are used to define the natural logarithm function as the inverse of ex. This follow naturally from the work on inverses. However, integration is involved and this is best saved until later. I will mention it then.

Two new post coming soon:

# Continuity

Karl Weierstrass (1815 – 1897) was the mathematician who (finally) formalized the definition of continuity. Included in that definition was the epsilon-delta definition of limit. This definition has been pulled out, so to speak, and now is usually presented on its own. So, which came first – continuity or limit? The ideas and situations that required continuity could only be formalized with the concept of limit. So, looking at functions that are or are not continuous helps us understand what limits are and why we first need them.

In the ideal world, students would have plenty of work with continuous and not continuous functions before starting the calculus. The vocabulary and notation, if not the formal definitions, would be used as early as possible. Then when students got to calculus, they would know the ideas and be ready to formalize the ideas.

The Intermediate Value Theorem (IVT) is an important property of continuous functions.

Using the definition of continuity to show that a function is or is not continuous at a point is a common question of the AP exams, as is the IVT.

Continuity The definition of continuity.

Continuity Should continuity come before limits?

From One Side or the Other One-sided limits and one-sided differentiability

How to Tell Your Asymptote from a Hole in the Graph  From the technology series. Showing holes and asymptotes on a graphing calculator.

Fun with Continuity Defined everywhere and continuous nowhere. Continuous only at a single point.

Theorems The Intermediate Value Theorem (IVT) and suggestions on teaching theorems.

Intermediate Weather  Using the IVT

Right Answer – Wrong Question Continuity or continuity “on its domain”?

Revised from a post of August 22, 2017

# The Mean Value Theorem

Another application of the derivative is the Mean Value Theorem (MVT). This theorem is very important. One of its most important uses is in proving the Fundamental Theorem of Calculus (FTC), which comes a little later in the year. Here are some previous post on the MVT:

Fermat’s Penultimate Theorem   A lemma for Rolle’s Theorem: Any function extreme value(s) on an open interval must occur where the derivative is zero or undefined.

Rolle’s Theorem   A lemma for the MVT: On an interval if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b) and f(a) = f(b), there must exist a number in the open interval (a, b) where ‘(c) = 0.

Mean Value Theorem I   Proof

Mean Value Theorem II   Graphical Considerations

Darboux’s Theorem   The Intermediate Value Theorem for derivatives.

Mean Tables

# Teaching and Learning Theorems

Theorems are carefully worded statements about mathematical facts that have been proved to be true. Important (and some not so important) ideas in calculus and all of mathematics are summarized as theorems. When you come across a theorem you need to understand it; the author of your textbook would not have included it and the AP Exams would not test it if it were not.  This post discusses some things about theorems in general. Students often do not realize these things; understanding them will help students understand a new theorem when it is presented to them.

Theorems have the form of IF one or more things are true (called the hypothesis), THEN some other thing is true (called the conclusion).  This is abbreviated $p\to q$ where p represents the hypothesis and q represents the conclusion. This is read as “if p, then q or “p implies q.” Theorems are also known as conditional statements.

In a certain instance, once you are sure all the conditions of the hypothesis are true, then you may be absolutely certain the conclusion is also true. When trying to determine if something is true about some function, check to see if the conditions of the hypothesis are all true.

People using a theorem need to know the hypotheses as well as the conclusion.

When teaching about a particular theorem, one thing that is often helpful to students is to “play” with the hypothesis and see how it affects the conclusion. For instance, if the hypothesis requires a function to be continuous, see what happens if the function is not continuous. If there are several parts to the hypothesis, see what happens if one or the other is changed. Hint: A change in part of the hypothesis will make some difference – good theorems do not have extra, unneeded, or superfluous conditions.

To make them read better, some theorems are not stated in if …, then… form. If there is any confusion, restate the theorem in if…, then… form.

• The theorem often stated as, “Differentiability implies continuity,” really means: IF a function is differentiable at a point, THEN it is continuous at that point.
• The geometry theorem, “The diagonals of a rhombus are perpendicular,” really means: IF a quadrilateral is a rhombus, THEN its diagonals are perpendicular.

The Contrapositive

Since if p is true, q must be true, what happens if q is false? The answer is that p must also be false. This is a related conditional statement (theorem) called the contrapositive of the theorem. The contrapositive is abbreviated IF not q, THEN not p, or IF q is false, THEN p is false, or $\tilde{\ }q\to \tilde{\ }p$.  The contrapositive of a theorem is also true, always.

There are several theorems in the calculus where the contrapositive seems to be used more often than the theorem itself.

• Differentiability implies continuity. The contrapositive of this theorem is: IF a function is not continuous at a point, THEN it is not differentiable at that point. This is a quick way to tell if a function is not differentiable.
• IF an infinite series converges, THEN the limit as n goes to infinity of its nth term is zero. Here the contrapositive is IF the limit as n goes to infinity of its nth term of an infinite series is not zero, THEN the series does not converge. This is called the nth term test for divergence.
• IF the diagonals of a quadrilateral are not perpendicular, THEN the quadrilateral is not a rhombus.

The converse

The converse of a theorem is a statement formed by switching the hypothesis and conclusion of the theorem: IF q, THEN p (or $q\to p$). The converse is not necessarily true. It may be true, in which case it need to be proved as a separate theorem. Students (among others) often assume that the converse is true – this is called the fallacy of the converse.

• IF a function is continuous at a point, THEN it is differentiable there, is the converse of our previous example. It is false: a simple counterexample is f(x) = |x|. This function is continuous at the origin but not differentiable there.
• Another theorem states that IF the derivative of a function is positive on an interval, THEN the function is increasing on the interval. The converse is, IF a function is increasing on an interval, THEN its derivative is positive on the interval. The converse is false: for example, f(x) = x3 is increasing everywhere, yet its derivative at the origin is zero. (See Going up?)
• IF the diagonals of a quadrilateral are perpendicular, Then the quadrilateral is a rhombus, if false. The quadrilateral may have perpendicular diagonals, but unless they intersect at their midpoints the figure is not a rhombus (it is a kite shape).

The inverse

The inverse is the contrapositive of the converse or IF not p, THEN not q (or$\tilde{\ }p\to \tilde{\ }q$ ). The inverse will be true if the converse is true, and false if the converse is false.

• The inverse of our fist example is,IF a function is not differentiable on an interval, THEN it is not continuous there. This is false, since the function my fail to be differentiable even though it is continuous. An example is f(x) = |x|. again.
• IF a quadrilateral is not a rhombus, THEN he diagonals are not perpendicular (false – the kite again).

Biconditional statements

A biconditional theorem is a theorem whose converse is also true (and therefore its inverse and contrapositive are true). These are written in the form p if, and only if, q or $p\leftrightarrow q$ (or for that matter $q\leftrightarrow p$). It is equivalent to $p\to q\text{ and }q\to p$.

• An example from Geometry: IF two sides of a triangle are congruent, THEN the angles opposite them are congruent. The converse of this theorem is, if two angles of a triangle are congruent,THEN the sides opposite them are congruent. Since both the theorem and its converse (and its inverse, and its contrapositive) are true, you may write, “Two sides of a triangle are congruent if, and only if, the angles opposite them are congruent.”

Definitions are always biconditional statements. They are always true and do not need to be proved; in fact they cannot be proved.

• The definition of continuous at a point is, “A function is continuous at a point $\left( {a,f\left( a \right)} \right)$ if, and only if, $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right)$ and both the limit and value are finite.”
• A rectangle is defined as a quadrilateral with four right angles or A quadrilateral is a rectangle if, and only if, it has four right angles.

Which is which?

The theorem, its contrapositive, converse, and inverse are all theorems. Any of them could be taken as “the theorem” and the others would rearrange their names accordingly. (Which is good practice for your students.)

One other thing

What if the hypothesis of a theorem is false: can the conclusion still be true? The answer is, yes! The hypothesis of a theorem tells us that if true, the conclusion must be true. But the conclusion may be true anyway.

Consider the Mean Vale Theorem (MVT): IF a function, f, is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), THEN there exists at least one number c in the open interval (a, b) such that $\displaystyle {f}'\left( c \right)=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$.

• But consider the function $f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}} & {-2\le x\le 1} \\ \text{4} & {1
• This function has a jump discontinuity at x = 1 and therefore is neither continuous nor differentiable on the interval [-2, 2]; the MVT does not apply. Yet $\displaystyle {f}'\left( 0 \right)=\frac{{f\left( 2 \right)-f\left( {-2} \right)}}{{2-\left( {-2} \right)}}=\frac{{4-4}}{4}=0$. (In fact, c could be 0 or any number between 1 and 2). In this example and not every example, the conclusion of the MVT is true even though the hypothesis is false.
• On the 2017 International exam AB 15 makes use of the idea that even though the theorem about limits that seems to apply doesn’t because the conditions are not met, but, nevertheless, the conclusion is true.

Proof

The proofs of all the important theorems are given in any good textbook. You study proofs for two reasons: (1) to see why a theorem is true, and (2) to learn how to write a proof of your own. If neither of these reasons concern you or your students (and they may not), then you still need to learn the hypothesis and conclusion, and how to apply the theorem. This cannot be avoided.

Some proofs are rather tricky. That is, the key step is not obvious. A beginning calculus student should not expect to know how to prove most of the theorems; they should, however, be able to follow the proofs in the textbook. The AP Calculus exams never ask for proof, per se, although they may ask you to justify a conclusion you make. The justification should show that the hypotheses are all true and state the name of the theorem that implies your conclusion.

I can recall only one time many years ago where students were asked to “prove” something on an AP Calculus Exam. The usual instruction is “Justify your answer” or “Explain your reasoning.” This means that students are supposed to cite the appropriate theorem and show that the hypotheses are met in the given situation. So, not quite “prove” but close. It’s not “prove” an original theorem, but rather determine which (unnamed) theorem applies (or does not apply) in a particular situation and verify that the conditions are (or are not) met.

As always, look at a number of past exams and see just what is asked and how it is asked.