Category Archives: Precalculus

What is an angle?

The notion of an angle is so basic (and we use it everyday indeed) that anyone with basic math knowledge would know what it is, right? Turns out it is not the case. I often was surprised to find out that even math majors in upper level math courses can’t explain what an angle is. They commonly answer like this: First draw two rays that intersect at a point and then draw an arc between them. They would call this arc an angle.

Figure 1. An angle

But then there can be so many different choices of arcs between two such rays. So which arc is an angle? The arc that defines an angle is one that is a part of the unit circle.

Figure 2. An angle

A measurement of an angle requires a unit. It is speculated that the degree measurement was invented by ancient Sumerians. Sumer is the earliest known civilization which was in southern Mesopotamia (modern day southern Iraq) and it dates back before 3,000 B.C. The biblical Genesis appears to have originated from Sumerian creation myth Enûma Eliš which predates Torah (Hebrew Bible). The first man’s name is Adamu and it does have the story of the Great Deluge. Sumerians had own writing system using cuneiform (wedge-shaped marks on clay tablets) and they built massive structures called ziggurats. Amazingly, those ziggurats still remain to this day in modern day Iraq. Sumerians possessed highly advanced level of knowledge in math and science including astronomy. Here is how they introduced the degree unit of angle measurement: Divide the unit circle into 360 equal sections (imagine 360 equal pizza slices). The length of the arc of each section is then defined to be $1^\circ$ and the circumference of the unit circle is $360^\circ$.

Figure 3. Degree angle measurement

Why the number 360? A common speculation is that Sumerians thought 1 year = 360 days, a complete rotation. Oh yes, they appear to have known that Earth is rotating around the Sun on a circular (actually elliptic) orbit. I have a different theory though. I do believe that Sumerians actually knew 1 year = 365 days. But if they used the number 365 to define $1^\circ$, it would have been an awfully ugly and inconvenient angle measurement. I believe that they knew this and used 360 instead.

On the other hand, you know from elementary geometry that the circumference of a circle with radius $r$ is $2\pi r$. So the unit circle has circumference $2\pi$. Hence we must have $$2\pi=360^\circ$$ However, something looks really awkward. The right hand side has a unit while the left hand side doesn’t. So it was given a unit called radian (denoted by rad), i.e. $$2\pi\ \mathrm{rad}=360^\circ$$ or $$\pi\ \mathrm{rad}=180^\circ$$ Note that 1 rad is merely a number (it’s same as the number 1) while $1^\circ$ is not. So in calculus we don’t use degree angle measurement but only use radian angle measurement.

Given an angle $\theta$, there is a corresponding point $(x,y)$ on the unit circle as shown in Figure 4.

Figure 4. Sine and cosine by unit circle

In order to make this correspondence more transparent, one may want to represent $x$ and $y$ in terms of the angle $\theta$. This is how cosine and sine of an angle were introduced: \begin{equation}\label{eq:trig}x=\cos\theta,\ y=\sin\theta\end{equation} The unit circle centered at the origin satisfies the equation \begin{equation}\label{eq:circ}x^2+y^2=1\end{equation} Using \eqref{eq:trig}, \eqref{eq:circ} can be written as \begin{equation}\label{eq:circ2}(\cos\theta)^2+(\sin\theta)^2=1\end{equation} But this looks a bit ugly, so a better looking notations were introduced: $$\cos^2\theta:=(\cos\theta)^2,\ \sin^2\theta:=(\sin\theta)^2$$ Thereby, \eqref{eq:circ2} is written as $$\cos^2\theta+\sin^2\theta=1$$ For an obvious reason, $\cos\theta$ and $\sin\theta$ along with $\tan\theta$, $\cot\theta$, $\sec\theta$, and $\csc\theta$ are called circular functions. I bet though most of you who are reading this have never heard of the name circular functions. The reason is that nowadays those quantities are defined more commonly (and shamefully) by using right triangles. (See Figure 5.) Hence they are called by more familiar name, trigonometic functions.

Figure 5. Sine and cosine by right triangle

An ancient Babylonian clay tablet which dates to 3,700 years ago, around the time when King Hammurabi ruled the Babylonian Empire, was recently decoded by scientists. It turns out that the tablet contains the oldest trigonometric table. In general, trigonometry is thought to have originated by Greek astronomer Hipparchu around 120 B.C. This remarkable discovery tells that ancient Babylonians already have known trigonometry 1,000 years before Greeks did. More details about the clay tablet can be read here.

Zeros of Polynomials

As we studied here, once you know how to find at least one rational zero of a polynomial using long division or synthetic division you can find the rest of the zeros of the polynomial. In this note, we study how to find a rational zero of a polynomial if there is one. Let $P(x)=a_nx^n+\cdots +a_1x+a_0$ and suppose that $P(x)$ has a rational zero $\frac{p}{q}$. This means that by factor theorem $P(x)$ has a factor $x-\frac{p}{q}$ or equivalently a factor $qx-p$. That is, $P(x)=(qx-p)Q(x)$ where $Q(x)$ is a polynomial of degree $n-1$. Let us write $Q(x)=b_{n-1}x^{n-1}+\cdots+b_1x+b_0$. Then we see that $a_n=qb_{n-1}$ and $a_0=-pb_0$. This means that $q$ is a factor of the leading coefficient $a_n$ of $P(x)$ and $p$ is a factor of the constant term $a_0$ of $P(x)$. Hence we have the Rational Zero Theorem.

Rational Zero Theorem. Let $P(x)=a_nx^n+\cdots +a_1x+a_0$ be a polynomial with integer coefficients where $a_n\ne 0$ and $a_0\ne 0$. If $P(x)$ has a rationa zero $\frac{p}{q}$ then $q$ is a factor of $a_n$ and $p$ is a factor of $a_0$.

Here is the strategy to find a rational zero of a polynomial $P(x)$.

STEP 1. Use the rational zero theorem to find the all candidates for a rational zero of $P(x)$.

STEP 2. Test each candidate from STEP 1 to see if it is a rational zero using the factor theorem. Once you find one say $\frac{p}{q}$, stop and move to STEP 3

STEP 3. Use long division or synthetic division (easier) to divide $P(x)$ by $x-\frac{p}{q}$ to find the rest of the zeros.

STEP 4. If necessary (in the event $Q(x)$ from STEP 3 has a higher degree), repeat the process $Q(x)$ from STEP 1.

Example. Find all zeros of $P(x)=2x^3+x^2-13x+6$.

Solution. $a_0=2$ has factors $\pm 1$ and $\pm 2$. $a_0=6$ has factors $\pm 1,\pm 2,\pm3\pm 6$. Thus all the candidates for a rational zero are
$$\pm 1,\pm 2,\pm 3,\pm 6,\pm\frac{1}{2},\pm\frac{2}{2}=\pm 1,\pm\frac{3}{2},\pm\frac{6}{2}=\pm 3$$
Since $P(2)=0$, 2 is a rational zero. Using long division or synthetic division we find $Q(x)=2x^2+5x-3=(2x-1)(x+3)$. Therefore, all zeros of $P(x)$ are $-3,\frac{1}{2},2$.

Example. Find all zeros of $P(x)=x^4-5x^3+23x+10$.

Solution. $a_n=1$ has factors $\pm 1$ and $a_0=10$ has factors $\pm 1, \pm 2, \pm 5, \pm10$. So all the candidates for a rational zero are
$$\pm 1, \pm 2, \pm 5, \pm10$$
Since $P(5)=0$, 5 is a rational zero. Using long division or synthetic division we find $Q(x)=x^3-5x-2$. We cannot factor this cubic polynomial readily so we repeat the process. The leading coeffient 1 has factors $\pm 1$ and the constant term $-2$ has factors $\pm 1,\pm 2$ so all the candidates for a rational zero of $Q(x)$ are $\pm 1,\pm 2$. $Q(-2)=0$ so $-2$ is a rational zero of $Q(x)$ (and hence of $P(x)$ as well). Using one’s favorite division we find the quotient $x^2-2x-1$ which has two real zeros $1\pm\sqrt{2}$. Therefore, all zeros of $P(x)$ are
$5, -2, 1\pm\sqrt{2}$.

It would be convenient if we can estimate how many positive real zeros and how many negative zeros without actually factoring the polynomial. Here is a machinary just for that.

Descartes’ Rule of Signs

Let $P(x)$ be a polynomial with real coefficients.

  1. The number of positive real zeros of $P(x)$ is either equal to the number of variations in sign in $P(x)$ or is less than that by an even number.
  2. The number of negative real zeros of $P(x)$ is either equal to the number of variations in sign in $P(-x)$ or is less than that by an even number.

Example. $P(x)=3x^6+4x^5+3x^3-x-3$ has one variation in sign so there is one positive real zero. $P(-x)=3x^6-4x^5-3x^3+x-3$ has three variations in sign so there can be either three negative zeros or one negative zero.

Upper and Lower Bounds for Real Zeros

Let $P(x)$ be a polynomial with real coefficients.

  1. If we divide $P(x)$ by $x-a$ ($a>0$) using synthetic division and if the row that contains the quotient and remainder has no negative entry, then $a$ is an upper bound for the real zeros of $P(x)$.
  2. If we divide $P(x)$ by $x-b$ ($b<0$) using synthetic division and if the row that contains the quotient and remainder has entries that are alternatively nonpositive and nonnegative, then $b$ is a lower bound for the real zeros of $P(x)$.

Example. If we divide $P(x)=3x^6+4x^5+3x^3-x-3$ by $x-1$ then
3 & 4 & 3 & 0 & -1 & -3\\
& 3 & 7 & 10 & 10 & 9\\
3 & 7 & 10 & 10 & 9 & 6
Since the row that contains quotient and remainder has no negative entries, 1 is an upper bound for real zeros of $P(x)$. If we divide $P(x)$ by $x-(-2)$ then
3 & 4 & 3 & 0 & -1 & -3\\
& -6 & 4 & -14 & 28 & -54\\
3 & -2 & 7 & -14 & 27 & -57
The entries of the row that contains the quotient and remainder are alternatively nonpositive and nonnegative, so $-2$ is a lower bound for real zeros of $P(x)$. $P(x)$ in fact does not have any integer zeros but the upper and lower bounds helps us graphically locate the real zeros of $P(x)$. Also they can be used as initial estimates for Newton’s method, a method that can find approximations to real zeros of a polynomial. Figure 1 shows that there is one positive real zero and one negative real zero of $P(x)$.

Real zeros of P(x)=3x^6+4x^5+3x^3-x-3

Dividing Polynomials

Polynomials are nice in the sense that they behave like numbers. For polynomials Division Algorithm works as well namely Given polynomials $P(x)$ and $D(x)\ne 0$ there exist unique polynomials $Q(x)$ and $R(x)$ such that
$P(x)$, $D(x)$, $Q(x)$, and $R(x)$ are called, respectively, the dividend, divisor, quptient and remainder. There are two ways to divide a polynomial by another polynomial. The first one is the familiar long division and it works the same way we do with numbers.

Example. Let $P(x)=8x^4+6x^2-3x+1$ and $D(x)=2x^2-x+2$. Find polynomials $Q(x)$ and $R(x)$ such that $P(x)=D(x)Q(x)+R(x)$.


Long division

Hence $Q(x)=4x^2+2x$ and $R(x)=-7x+1$.

The other method is called synthetic division. This method cannot be used for any polynomial divisions, however it works great when the divisor is a linear polynomial and is easier than long division. Synthetic division uses only coefficients without including variables as shown in the following example.

Example. Using synthetic division divide $2x^3-7x^2+5$ by $x-3$.


Synthetic division

Hence we have $Q(x)=2x^2-x-3$ and $R=-4$.

If a polynomial $P(x)$ is divided by a linear polynomial $x-c$, by division algorithm $P(x)$ can be written as
for some $Q(x)$ and $R$. So, $P(c)=R$ and hence we obtain the Remainder Theorem.

Remainder Theorem. If a polynomial $P(x)$ is divided by $x-c$, then the remainder is $P(c)$.

Example. Let $P(x)=3x^5+5x^4-4x^3+7x+3$. Use the remainder theorem to find the remainder when $P(x)$ is divided by $x+2$.

Solution. $R=P(-2)=5$.

As a corollary of the remainder theorem we have

Factor Theorem. $c$ is a zero of $P(x)$ if and only if $x-c$ is a factor of $P(x)$.

Example. Let $P(x)=x^3-7x+6$. Show that $P(1)=0$ and use this information to factor $P(x)$ completely.

Solution. Dividing $P(x)$ by $x-1$ (using long division or synthetic division) we find $Q(x)=x^2+x-6$ and $R=0$ (of course as we expected). So,

Example. Find a polynomial of degree 4 that has zeros $-3$, 0, 1, and 5.

Solution. Such a polynomial would have $x+3$, $x$, $x-1$, and $x-5$ for its factors by the factor theorem. So the simplest one is

One-to-One Functions and Inverse Functions

A function $y=f(x)$ is said to be one-to-one if satisfies the property
$$f(x_1)=f(x_2) \Longrightarrow x_1=x_2$$
or equivalently
$$x_1\ne x_2 \Longrightarrow f(x_1)\ne f(x_2)$$
for all $x_1,x_2$ in the domain. In plain English what this says is no two numbers in the domain are corresponded to the same number in the range. Figure 1 is the graph of $f(x)=x^2$. It is not one-to-one.

Figure 1. The graph of y=x^2

For example, $-1\ne 1$ but $f(-1)=1=f(1)$.

Figure 2. The graph of y=x^3

Figure 2 is the graph of $f(x)=x^3$. It is one-to-one as seen clearly from the graph. But let us pretend that we don’t know the graph but want to prove that it is one-to-one following the definition. Here we go. Suppose that $f(x_1)=f(x_2)$. Then $x_1^3=x_2^3$ or $x_1^3-x_2^3=(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0$. This means $x_1=x_2$ which completes the proof.

Why do we care about one-to-one functions? The reason is that if $y=f(x)$ is one-to-one, it has an inverse function $y=f^{-1}(x)$.
x&\stackrel{f}{\longrightarrow} y\\
x&\stackrel{f^{-1}}{\longleftarrow} y
Given a one-to-one function $y=f(x)$, here is how to find its inverse function $y=f^{-1}(x)$

STEP 1. Swap $x$ and $y$ in $y=f(x)$. The reason we are doing this is that $\mathrm{Dom}(f)=\mathrm{Range}(f^{-1})$ and $\mathrm{Dom}(f^{-1})=\mathrm{Range}(f)$.

STEP 2. Solve the resulting expression $x=f(y)$ for $y$. That is the inverse function $y=f^{-1}(x)$.

Example. Find the inverse function of $f(x)=\frac{2x+3}{x-1}$. (It is a one-to-one function.)

Solution. STEP 1. Let $y=\frac{2x+3}{x-1}$ and swap $x$ and $y$. Then we have

STEP 2. Let us solve $x=\frac{2y+3}{y-1}$ for $y$. First multiply $x=\frac{2y+3}{y-1}$ by $y-1$. Then we have $x(y-1)=2y+3$ or $xy-x=2y+3$. Isolating the terms that contain $y$ in the LHS, we get $xy-2y=x+3$ or $(x-2)y=x+3$. Finally we find $y=\frac{x+3}{x-2}$. This is the inverse function.

$y=f(x)$ and its inverse $y=f^{-1}(x)$ satisfy the following properties.
$$(f\circ f^{-1})(x)=x,\ (f^{-1}\circ f)(x)=x$$
The reason for these properties to hold is clear from the definition of an inverse function. We can check the properties using the above example. I will do $(f\circ f^{-1})(x)=x$ and leave the other for an exercise.
(f\circ f^{-1})(x)&=f(f^{-1}(x))\\
The graph of $y=f(x)$ and the graph of its inverse $y=f^{-1}(x)$ satisfy a nice symmetry, namely they are symmetric about the line $y=x$. This symmetry helps us obtain the graph of $y=f^{-1}(x)$ when the explicit expression for $f^{-1}(x)$ is not available. You will see such a case later when you study the logarithmic functions. Figure 3 shows the symmetry with $y=x^2$ ($x\geq 0$) and its inverse $y=\sqrt(x)$.

Figure 3. The symmetry between the graphs of y=x^2 (red) and y=sqrt(x) (blue) about y=x (green)

Combining Functions

It’s quite interesting that functions can be treated like numbers, namely you can define $+$, $-$, $\times$, and $\div$ on a collection of functions. How do we do this? For instance given two functions $f$ and $g$, we can define a new function $f+g$ by
for all $x$ in the domain (for sake of simplicity we assume that $f$ and $g$ both have the same domain. If not, one can take the intersection of the domains of $f$ and $g$, no big deal). In a similar manner, we can also define $f-g$, $fg$, and $\frac{f}{g}$ respectively as
\left(\frac{f}{g}\right)(x)&=\frac{f(x)}{g(x)}\ \mbox{provided}\ g(x)\ne 0

Example. Let $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$.

(a) Find the functions $f+g$, $f-g$, $fg$ and $\frac{f}{g}$ and their domains.

(b) Find $(f+g)(4)$, $(f-g)(4)$, $(fg)(4)$, $\left(\frac{f}{g}\right)(4)$

Solution. (a) $\mathrm{Dom}(f)=\{x|x\ne 2\}$ and $\mathrm{Dom}(g)=\{x|x\geq 0\}$. So the intersection is $\{x|0\leq x<2\}\cup\{x|x>2\}=[0,2)\cup(2,\infty)$ and this is the domain of $f+g$, $f-g$ and $fg$. For $\frac{f}{g}$ since $g$ is not defined at $x=0$, its domain should be $(0,2)\cup(2,\infty)$.

(b) I will do only $(f+g)(4)$. One way to evaluate $(f+g)(4)$ is to use $(f+g)(x)$ we obtained in part (a) i.e. $(f+g)(4)=\frac{1}{4-2}+\sqrt{4}=\frac{5}{2}$. Another way is evaluating $f(4)$ and $g(4)$ first which are $f(4)=\frac{1}{2}$ and $g(4)=2$. Then $(f+g)(4)=f(4)+g(4)=\frac{5}{2}$.

Composite Functions

Given two functions $f$ and $g$, if the range of $f$ is a subset of the domain of $g$, then we can combine the two functions to create a new function which we will denote by $g\circ f$.
$$x\stackrel{f}{\longmapsto} f(x)\stackrel{g}{\longmapsto} g(f(x))$$
The above diagram hints us that we can define a new function $g\circ f$ by
$$(g\circ f)(x)=g(f(x))$$
We call $g\circ f$ “$f$ followed by $g$.”

Example. Let $f(x)=x^2$ and $g(x)=x-3$.

(a) Find composite functions $f\circ g$ and $g\circ f$ and their domains.

(b) Find $(f\circ g)(5)$ and $(g\circ f)(5)$.

Solution. (a) By definition $(f\circ g)(x)=f(g(x))=f(x-3)=(x-3)^2$. Also by definition $(g\circ f)(x)=g(f(x))=g(x^2)=x^2-3$. From these we can clearly see both their domains are $(-\infty,\infty)$. In general $\mathrm{Dom}(f\circ g)=\mathrm{Dom}(g)$ and $\mathrm{Dom}(g\circ f)=\mathrm{Dom}(f)$.

(b) $(f\circ g)(5)$ can be evaluated using $(f\circ g)(x)$ we obtained in part (a).
$$(f\circ g)(5)=(5-3)^2=4$$
There is another way to do this. If you don’t have to find $(f\circ g)(x)$ but only need to calculate $(f\circ g)(5)$, this may be simpler. First note $(f\circ g)(5)=f(g(5))$. $g(5)=5-3=2$, so $f(g(5))=f(2)=2^2=4$. Similarly we find $(g\circ f)(5)=22$. In general $(f\circ g)(x)\ne (g\circ f)(x)$.