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Math 31A - Review of Precalculus
1
Introduction
One frequent complaint voiced by reluctant students of mathematics is: ‘Why do I have to study this? When
will I need math in real life?’ While it is my personal belief that mathematics is a beautiful subject that
should be studied (worshipped?) for aesthetic reasons alone, a simpler answer to the question is that mathematics is the language through which we express our understanding of the physical world. (This contentious
claim may raise some interesting philosophical questions, but they are beyond the scope of this course.)
In most numerate disciplines, our knowledge is encapsulated in mathematical models, which more often
than not involve calculus. Calculus is one of the most widely applied areas of mathematics, because it allows
us to examine the relationship between different variables. For instance, in kinetics we are interested in
studying velocity - the rate of change of position with time - and acceleration - the rate of change of velocity
with time. Calculus makes this possible.
While some of the ideas of calculus have been around for thousands of years, the formal theory was
developed (independently) by Newton and Leibniz in the 1600’s. This course provides an introduction to
the subject, which will allow you to apply the concepts to problems you will encounter in other classes (and
in the Real World). However, it is necessary to have a firm grasp of the prerequisites (usually covered in
a precalculus course), and these notes are intended to act as a review of functions and graphs, which will
appear throughout the quarter.
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Functions
Just as the operations of arithmetic (addition, subtraction, multiplication, division, etc) are applied to
numbers, the operations of calculus (differentiation and integration) are applied to functions. Hence a firm
understanding of functions is required before we can begin our study of calculus.
2.1
Definition of a function
You can think of a function as a machine that accepts an input value and returns an output value. For a
function f and an input x, we denote the output as f (x). It is important that for every input value, there is
precisely one output value.
Figure 1: A function as a machine
While a function can be defined more generally, we will typically consider functions whose input and
output values are real numbers.
2.2
Key Terminology
The following are some definitions of terminology related to functions:
The domain of a function is the set of input values for which the function is defined. For example, the
domain of the function f (x) = x1 is the set of non-zero real numbers; we cannot divide by 0, so 0 is not in
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the domain of f .
The range of a function is the set of possible output values of the function (when applied to its domain).
For instance, the range of the function g(x) = x2 is the set of non-negative numbers, as x2 ≥ 0 for any value x.
The following table gives some examples of domains and ranges of functions. See if you can fill in the
blanks. (R represents the set of all real numbers.)
f (x)
√
x
cos x
1
2x−1
Domain
{x : x ≥ 0}
R
{x : x 6= 21 }
Range
{y : y ≥ 0}
{y : −1 ≤ y ≤ 1}
{y : y 6= 0}
tan x
√
x2 − 2
The independent variable is the input of the function, usually denoted by x. If writing an equation
in the form y = f (x), y is called the dependent variable because its value depends on the independent
variable x.
A zero (or root) of a function f is an input value x such that the corresponding output value is zero, or
f (x) = 0.
2.3
Graphs of Functions
While the algebraic form of the functions presented above is useful for computing exact values, it does not
give us a clear picture of the overall behaviour of the function. An alternate way to present a function is its
graph - this is a pictorial representation that illustrates global features very clearly. The graph of a function
is obtained by plotting the points (x, f (x)) on a coordinate plane. The x-axis represents the independent
variable (the input values), while the y-axis represents the dependent variable (the output values). The height
of the function above/below the x-axis at a point x represents the value f (x) at that point. An example of
the graph of the function f (x) = 2x + 10 is given below.
Figure 2: Graph of the function f (x) = 2x + 10
Note that if we are presented with a curve in a plane, the Vertical Line Test allows us to tell whether
it is the graph of a function or not. The test says that every vertical line (x = a) should intersect the curve
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at most once. That is to say, if any vertical line intersects the curve twice or more, then the curve does not
represent a function. This is because the function can only have one output value for every input value.
2.4
Properties of Functions
As mentioned above, the graph of a function allows us to identify certain properties of the function very
quickly. We list some such properties below:
A function is increasing if larger input values result in larger output values. Formally, this expressed as:
if x2 ≥ x1 , then f (x2 ) ≥ f (x1 ). The function is strictly increasing if a slightly stronger condition holds: if
x2 > x1 , then f (x2 ) > f (x1 ). Graphically, this means the graph goes up as it goes to the right.
Similarly, a function is decreasing if larger input values result in smaller output values. More formally, if
x2 ≥ x1 , then f (x2 ) ≤ f (x1 ). As you may suspect, a function is said to be strictly decreasing if whenever
x2 > x1 , then f (x2 ) < f (x1 ). Graphically, the graph goes down as it goes to the right.
A function is even if f (−x) = f (x). This means the graph is symmetric with respect to the y-axis.
A function is odd if f (−x) = −f (x). This means the graph is symmetric with respect to the origin (i.e.
the line joining (x, f (x)) and (−x, f (−x)) passes through the origin).
Thus the first two properties determine the direction of the graph, while the latter two determine the
symmetry of the graph. These properties are useful if trying to sketch the graph of a given function.
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Polynomials
There are many, many functions (infinitely many, and a very big type of infinity), and some functions are
nicer than others. Perhaps the nicest class of (interesting) functions is the polynomials. These functions
are sums of multiples of non-negative integer powers of x. For example, f (x) = 3.54x19 + 2x12 + 0.12391x + 21
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is a polynomial (the constant term is valid as x0 = 1), while f (x) = x 2 is not a polynomial, since the power
is not a non-negative integer. The degree of a polynomial is the highest power of x that appears (with a
non-zero coefficient). Thus the degree of the polynomial in the example is 19.
There are many reasons why polynomials are nice functions (perhaps even infinitely many), which include
the fact that they are not only continuous but can be differentiated infinitely many times, and that any
continuous function can be approximated by polynomials (with an error as small as you want). Most of these
reasons will have to wait until a course on analysis, but in this course we will be able to appreciate some of
the nice calculus-related properties they have.
3.1
Linear Polynomials
The simplest polynomials are the linear polynomials, which are polynomials of degree at most one. The name
comes from the fact that linear polynomials represent straight lines.
3.1.1
Equations
As these are polynomials of degree at most one, they can be represented in the standard form
f (x) = ax + b
where a and b are real numbers. This is sometimes called the slope-intercept form of the equation of the
line. This is because the quantity a represents the slope of the line y = f (x) (the increase in y per unit
increase in x), while the quantity b is the y-intercept (the value where the line crosses the y-axis).
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There are a couple of alternative forms of the equation of a line. If we know that the slope of the line is
m and the line passes through the point (x0 , y0 ), then we can express the line as
y − y0 = m(x − x0 )
This is the point-slope form of the equation.
Alternatively, if we know the line passes through the points (x1 , y1 ) and (x2 , y2 ), then the two-point
equation for the line is
y2 − y1
y − y1 =
(x − x1 )
x2 − x1
One can easily switch between the different forms with a little bit of algebraic manipulation.
3.1.2
Graphs
As stated above, linear polynomials represent straight lines. In particular, every non-vertical straight line in
the plane is the graph of a linear polynomial (vertical lines, of course, fail the vertical line test, and hence
are not functions). Below we show the graphs of some sample linear polynomials:
The function y = 3
The function y = 1.3x − 4
The function y = −3x + 2
If a = 0, then we get a horizontal line. If a 6= 0, then we get a non-horizontal line; if a > 0, it is
increasing, while if a < 0, it is decreasing. In particular, this line must have a unique root. We can solve for
it algebraically: if f (x) = ax + b = 0, then x = −b
a .
To plot the graph of a linear polynomial, just find two points on the line and plot the unique straight
line between them. These two points can be obtained by substituting any two input values x. Letting x = 0
gives the y-intercept, and can often be a useful point for plotting.
3.2
Quadratic Polynomials
While the simplicity of linear polynomials may make them a favourite of students, unfortunately it also means
that they are not complex enough to be used in many situations. Thus we need also consider polynomials of
degree two, or quadratic functions.
3.2.1
Equations of Quadratics
As a polynomial of degree two, quadratics have the standard form
f (x) = ax2 + bx + c
where a 6= 0 (otherwise it would be linear). As before, some algebra can give us different forms of the equation.
Another useful form of the quadratic is the vertex form, where we write
f (x) = A(x − x0 )2 + y0
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The point (x0 , y0 ) is called the vertex, and as we shall see when we plot quadratics, this is the turning point
of the graph.
To see the relationship between the coefficients of the standard form and the vertex form, we expand the
vertex form to get
f (x) = A(x − x0 )2 + y0 = A(x2 − 2x0 x + x20 ) + y0 = Ax2 − 2Ax0 x + (Ax20 + y0 )
Comparing this to the standard form f (x) = ax2 + bx + c, we find that a = A, b = −2Ax0 and c = (Ax20 + y0 ).
To get the vertex form from the standard form, we solve for A, x0 and y0 in terms of a, b and c to get A = a,
b2
x0 = −b
2a and y0 = c − 4a . The process of converting the standard form into the vertex form is known as
‘completing the square’.
Finally, when the quadratic has real roots (more on that in the next section), we can also write the
equation in factored form, namely
f (x) = A(x − x1 )(x − x2 )
Here A is the coefficient of the x2 term (so A = a), and x1 and x2 are the roots of the quadratic (so
f (x1 ) = f (x2 ) = 0).
3.2.2
Roots of Quadratics
Not every quadratic has a (real) root (and thus not every quadratic can be written in factored form). For
instance, if f (x) = x2 + 1, then since x2 ≥ 0 for every x, we have f (x) ≥ 1 for every x, and hence f (x) is
never zero.
The quadratic formula tells us exactly when a given quadratic has roots, and what they are (if they
exist). According to the formula, the roots of the quadratic f (x) = ax2 + bx + c are
√
−b ± b2 − 4ac
2a
Note that this expression is only defined when the term inside the square root, b2 − 4ac, is non-negative.
This term has a special name; it is called the discriminant, and is usually denoted by ∆ = b2 − 4ac.
In summary, given a quadratic in standard form, we wish to determine whether it has roots, and if so;
what they are. We first compute the discriminant, and the following table shows the possible cases:
Discriminant
∆<0
∆=0
∆>0
No. of distinct roots
0
1
2
Roots
none
−b
2a√
−b± ∆
2a
If the quadratic is not given in standard form, then determination of the roots is even easier. In vertex
form (f (x) = A(x − x0 )2 + y0 ), we have
y0
No. of distinct roots
Ay0 < 0
2
Roots
q
0
x0 ± −y
A
y0 = 0
Ay0 > 0
1
0
x0
none
If the quadratic is given in factored form (f (x) = A(x − x1 )(x − x2 )), then the roots are given to us
explicitly; they are x = x1 and x = x2 .
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3.2.3
Graphs of Quadratics
The shape of the graph of a quadratic function should be a familiar shape - it is a parabola! It can come in
one of two varieties:
‘Happy parabola’ : a > 0
‘Sad parabola’ : a < 0
Note that the turning point of the parabola (its local minimum or maximum) is located at the vertex as
defined in the vertex form of the equation. Also, if the parabola has its vertex at (x0 , y0 ), then it is symmetric
through the line x = x0 .
Thus, as described in the preceding sections, from the equation of the parabola we can extract the following
information:
• Whether it is upwards or downwards (a > 0 or a < 0)
• Where its vertex is
• How many roots it has, and where they are
This is enough for us to be able to sketch a fairly accurate graph of the equation. Below are some more
examples of quadratics and their graphs:
f (x) = 2x2 + x − 1
f (x) = −x2 + 3x − 2
An upward parabola with two distinct roots
A downward parabola with two distinct roots
f (x) = 4x2 − 4x + 1
f (x) = 2x2 − 4x + 3
An upward parabola with one root
An upward parabola with no roots
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3.3
Cubic Polynomials
We end this section by moving one degree higher and looking at polynomials of degree three, or cubic
polynomials.
3.3.1
Equations and Roots of Cubics
As polynomials of degree three, cubic functions have the standard form
f (x) = ax3 + bx2 + cx + d
where a 6= 0.
Unlike quadratics, which could possibly have no real roots, cubic equations must always have at least one
real root. (In fact, it is generally true that any odd-degree polynomial must have at least one real root, while
even-degree polynomials need not have any roots. The maximum number of roots a polynomial of degree n
can have is n.)
Unfortunately, there is no counterpart of the quadratic formula for cubics; there is no nice formula for
the roots of a cubic polynomial. If you are asked to find the roots of a cubic polynomial, you are generally
in one of two situations:
• You are allowed to use a computer/graphing calculator, which have solvers that can find a root for you.
• You are expected to do this by hand, and there is an ’easy’ root - usually a small integer (e.g. 0, 1, -1,
2, -2) that you can find by trial and error.
Suppose you have now found a root, x = x1 , so that f (x1 ) = 0. Then you can use polynomial long
division to factor the cubic into a linear term times a quadratic term (there will be no remainder)
f (x) = ax3 + bx2 + cx + d = (x − x1 )(Ax2 + Bx + C)
If you would like a review of how polynomial long division works, try checking out
http://www.sosmath.com/algebra/factor/fac01/fac01.html
We can then apply our discriminant analysis and quadratic formula to determine whether the quadratic
factor has roots. If so, then these roots are also roots of the cubic polynomial.
For example, consider the polynomial f (x) = x3 + 3x2 − 7x + 3. By trial and error, we find f (1) =
1 + 3 − 7 + 3 = 0, and so 1 is a root of the cubic. This means that x − 1 is a factor of the cubic polynomial.
Indeed, if we carry out polynomial division, we find
f (x) = x3 + 3x2 − 7x + 3 = (x − 1)(x2 + 4x − 3)
Evaluating the quadratic factor x2 +4x−3, we see the discriminant is ∆ = 42 −4·1·(−3)
= 16+12 = 28 > 0.
√
√
Hence the quadratic has two distinct roots, which by the quadratic formula are −4±2 28 = −2 ± 7. Thus
√
√
the roots of the cubic polynomial are −2 − 7, −2 + 7, 1.
3.3.2
Graphs of Cubics
In a parabola, we saw that the behaviour of the function is the same as x goes to positive or negative infinity.
For instance, in the case of an upward parabola, if x is very large and positive, then f (x) is very large and
positive. If x is very large and negative, then f (x) is still very large and positive.
With cubics, the behaviour at the two ends of the real axis is always different. When the leading coefficient
is positive, a > 0, then f (x) is very large and positive for large positive x, while f (x) is very large and negative
for large negative x. If a < 0, then the relationship is inverted - f (x) is large and negative for large positive
x, and f (x) is large and positive for large negative x. The two examples below illustrate this property.
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f (x) = x3 + 3x2 − 4x + 2
f (x) = −x3 + 3x2 − 4x + 2
So whenever you have to sketch the graph of a cubic, always start with the leading coefficient - use it to
determine the direction of the cubic.
We also have to look out for roots of the cubic. The previous section explained how to find the roots of
the cubic - try to find one by trial and error, divide to get a quadratic, and then use the discriminant and
quadratic formula to find the remaining roots. Depending on the form of the quadratic factor, you can have 1,
2 or 3 different roots (but you must always have at least 1). The pictures below show the different possibilities.
Upward Cubic with One Root
Upward Cubic with Two Roots
Upward Cubic with Three Roots
f (x) = x3 − 3x2 + 3x − 1
f (x) = x3 + x2 − x − 1
f (x) = x3 + 2x2 − x − 2
Downward Cubic with One Root
Downward Cubic with Two Roots
Downward Cubic with Three Roots
f (x) = −x3 + 3x2 − 3x + 1
f (x) = −x3 − x2 + x + 1
f (x) = −x3 − 2x2 + x + 2
Note that when there are multiple roots, there should be a turning point between every pair of consecutive
roots. This should help guide your sketch.
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4
Composition of Functions
Most functions are very complicated, and direct analysis would be very messy. One way to make them easier
to study is to build them up through composition of simpler functions.
4.1
Definition of composed functions
Recall that we can think of functions as machines - we input one number and get another number as the
output. Composition of functions is stringing these machines together (in series), so that the output of one
machine becomes the input of another. This is analogous to a manufacturing line; a series of machines work
together, with each completing a simple task before passing the result onto the next machine. In this way,
complex products can be built one simple step at a time.
Figure 3: Composition is Linking the Function Machines Together
Suppose we have two functions, f (x) and g(x). The composition f ◦ g (read ‘f composed with g’)
is the function that results when we apply g to x, and then feed the output g(x) as input to f . Thus
(f ◦ g)(x) = f (g(x)). Note that the order the functions are applied is inside-out, or right-to-left in the
composition notation.
4.2
Evaluation of compositions
Given f and g, we get an expression for f ◦ g by simply replacing x in the equation for f with g(x). Then
we can substitute in the equation for g(x) (remember to use parenthesis - constants should factor through).
For instance, suppose f (x) = x2 + 2x + 1 and g(x) = x − 3. Then we have
(f ◦ g)(x) = f (g(x)) = (g(x))2 + 2(g(x)) + 1 = (x − 3)2 + 2(x − 3) + 1
If necessary, we could expand further to get (f ◦ g)(x) = x2 − 4x + 4.
This is a good place to note that the order of the composition matters! If we were to instead evaluate
g ◦ f , then we get
(g ◦ f )(x) = g(f (x)) = g(x2 + 2x + 1) = (x2 + 2x + 1) − 3 = x2 + 2x − 2 6= (f ◦ g)(x)
We can repeat the process for any number of functions. If f (x) = x2 , g(x) = x + 1 and h(x) =
√
√
√
√
(f ◦ g ◦ h)(x) = f (g(h(x))) = f (g( x)) = f (( x) + 1) = (( x) + 1)2 = x + 2 x + 1
√
x, then
Note, however, that if we are only interesting in particular values of the composition, then we can avoid
the algebraic
manipulation. In the previous example, if we wanted to find (f ◦ g ◦ h)(4), then we note that
√
h(4) = 4 = 2, g(h(4)) = g(2) = 2 + 1 = 3, and so (f ◦ g ◦ h)(4) = f (g(h(4))) = f (3) = 32 = 9.
4.3
Decomposing Functions
It is often useful to be able to decompose functions as well - given a complicated function, we want to write it
as the composition of simpler functions. This is useful because it can separate the complexity of the original
function into separate parts, which we can then deal with later (you will see this in action when we learn the
chain rule).
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√
This is perhaps easiest to see by example. Suppose we have the function f (x) = 1 − cos3 x. There are
two complicated
expressions here - the square root and the cosine cubed. If we parenthesise the cosine cubed
p
(f (x) = √1 − (cos3 x)), this suggests letting the inner function be h(x) = cos3 x, and then the outer function
is g(x) = 1 − x. We then have f (x) = (g◦h)(x), reducing the function to a composition of simpler functions.
√
Note that this is not the only way to decompose the function. We could also have used G(x) = x and
H(x) = 1√− cos3 x, and then f (x) = (G ◦ H)(x). Moreover, we could have decomposed it into three functions:
f1 (x) = x, f2 (x) = 1 − x3 , and f3 (x) = cos x. Then f (x) = (f1 ◦ f2 ◦ f3 )(x). These three functions are even
simpler, but the trade-off is that we have more functions.
4.4
Domains and Ranges of Compositions
Recall that the domain of a function is the set of possible inputs, while the range is the set of possible
outputs. Determining the domain and range of a composed function can be a little tricky, since the domains
and ranges of the two functions interact a little. In my experience, it is easier to try and find the domain first.
√
f (g(x)). If x
For example, let f (x) = x and g(x) = 4 − x2 . We wish to find the domain of (f ◦ g)(x) = √
is in the domain, then f (g(x)) must make sense, so g(x) must be in the domain of f . f (x) = x, which is
defined for x ≥ 0. Hence for f (g(x)) to be defined, we must have g(x) = 4 − x2 ≥ 0. Solving this, we find
x2 ≤ 4, so −2 ≤ x ≤ 2. Hence the domain of (f ◦ g) is [−2, 2].
Now what about the range? We know from our above calculations that when x is in [−2, 2], g(x) ≥ 0.
However, g(x) = 4 − x2 , and x2 is always non-negative, so g(x) ≤ 4. Hence we are feeding the interval [0, 4]
√
into f . If 0 ≤ y ≤ 4, then 0 ≤ y ≤ 2, and so 0 ≤ f (g(x)) ≤ 2 for x in the domain [−2, 2]. So we see the
range is [0, 2].
To review, we first list the domain D of the outer function f . Then we have to find which values of x
can go into the inner function g such that the result g(x) lies in D. This gives the domain for the composed
function. To get the range of the composed function, we have to see what g does to the domain, and then
see what f does to that set.
If we have more than two functions being composed, we can just continue this process. Note that
F = f ◦ g ◦ h = f ◦ (g ◦ h), so if we let G = g ◦ h, then F = f ◦ G. We can use the method to find the range
and domain of G, and then use that information to find the range and domain of F .
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Conclusion
I realise these notes may be a little rushed, but they are intended to hopefully awaken your memories of the
material rather than teach it. Also, you may have been taught different methods for approaching these kind
of problems. Provided your method works and you understand why it works, that is perfectly fine. If you
have any questions about any of the above, please do e-mail me.
I wish you all the very best for this course. Have a great quarter!
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