MHF4U Advanced Functions – Course Notes
Unit 1 – Polynomial Functions and Equations
Def.: 1. A relation is a set of ordered pairs, ( x, y) , where x is the input (independent variable), and y is the
output (dependent variable).
2. The set of all inputs is called the domain, and the set of all outputs is called the range.
Def.: A function is a relation whereby each element of the domain is assigned to a unique element in the
range.
Representations of Functions
Note: A function can be represented explicitly (numerically, e.g. table of values, set of ordered pairs),
graphically or analytically/algebraically (equation).
Function Notation
Note: A function can be denoted using a letter, say f or g. An ordered pair (𝑥, 𝑦) is represented by 𝑦 = 𝑓(𝑥).
Piecewise Functions
Discussion: Consider a race car going around a track or the behaviour of the stock market. One equation may
not be sufficient to describe the behaviour, hence the need for piecewise functions.
Def.: A piecewise function is a function which is defined using more than one equation over different
intervals of the domain.
Ex.:
y = x can be defined as a piecewise function: 𝑦 = {
𝑥, 𝑖𝑓 𝑥 ≥ 0
−𝑥, 𝑖𝑓 𝑥 < 0
𝑥2,
0≤𝑥≤2
𝑓(𝑥) = { 𝑥 + 2, 2 < 𝑥 < 4
−𝑥 + 8,
𝑥≥4
Interval Notation
Def.: 1. [𝑥1 , 𝑥2 ] represents the interval from 𝑥1 to 𝑥2 , including the endpoints, i.e. 𝑥1 ≤ 𝑥 ≤ 𝑥2 .
2. (𝑥1 , 𝑥2 ) represents the interval from 𝑥1 to 𝑥2 , not including the endpoints, i.e. 𝑥1 < 𝑥 < 𝑥2 .
3. Intervals can possess both round and square brackets to indicate the inclusion or exclusion of an
endpoint.
4. If an interval increases to positive infinity (+∞) or decreases to negative infinity (−∞), a round
bracket must be used.
5. We use the “union symbol” ∪ to join two separate intervals
e.g. 𝑥 ≤ 0 or 𝑥 > 10 can be written (−∞, 0] ∪ (10, +∞)
Polynomial Functions
Discussion: What do we mean by a polynomial expression? Use this to define the idea of a polynomial
function.
Def.: A polynomial function of degree n is a function which takes the form
𝑓(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + 𝑎𝑛−2 𝑥 𝑛−2 + ⋯ + 𝑎2 𝑥 2 + 𝑎1 𝑥 + 𝑎0
Where,
1. 𝑛 ∈ ℕ
2. 𝑎𝑛 , 𝑛 = 0, 1, … , 𝑛 are called the coefficients.
3. 𝑎𝑛 ≠ 0 is called the leading coefficient.
4. 𝑎0 is called the constant coefficient.
Ex.: Linear and quadratic functions are polynomials. (Despite #3, y = 0 is a polynomial)
Def.: A function of the form 𝑓(𝑥) = 𝑎𝑥 𝑛 , where 𝑛 ∈ ℝ, is called a power function. Note that a power
function is a polynomial only if 𝑛 ∈ ℕ.
Def.: 1. A function 𝑓 has a local (relative) maximum at 𝑥 = 𝑎 if 𝑓(𝑎) ≥ 𝑓(𝑥) for all x in a neighbourhood
of a.
2. A function 𝑓 has a local (relative) minimum at 𝑥 = 𝑎 if 𝑓(𝑎) ≤ 𝑓(𝑥) for all x in a neighbourhood
of a.
3. A function f has an absolute maximum at 𝑥 = 𝑎 if 𝑓(𝑎) ≥ 𝑓(𝑥) for all x in the domain.
4. A function f has an absolute minimum at 𝑥 = 𝑎 if 𝑓(𝑎) ≤ 𝑓(𝑥) for all x in the domain.
5. In general, maxima and minima are referred to as extrema.
6. The end behaviour of a function refers to the behaviour of the y values as x approaches positive and
negative infinity (the behaviour need not be the same on both sides).
Examples:
1. 𝑦 = 𝑥 2 has both a local minimum and an absolute minimum at 𝑥 = 0.
2. 𝑦 = 2𝑥 has no local and no absolute extrema.
1 𝑥
3. End behaviour of 𝑦 = (2) : as 𝑥 → +∞, 𝑦 → 0 and as 𝑥 → −∞, 𝑦 → +∞.
Def.: A function is said to be continuous if “it can be drawn without lifting the pencil from the paper”, that is,
it has no gaps or jumps. (A more formal definition will come later)
Note: Properties of polynomials. Let f be a polynomial of degree n.
1. All polynomials
• Domain is ℝ.
• Have at most n x-intercepts.
• 𝑎𝑜 is the y-intercept.
• Continuous on ℝ.
• Have at most n − 1 extrema
2. Even-degree
• 𝑓(𝑥) → +∞ as 𝑥 → ±∞ if 𝑎𝑛 > 0
• 𝑓(𝑥) → −∞ as 𝑥 → ±∞ if 𝑎𝑛 < 0
• Range is restricted. (Must have an absolute extremum)
3. Odd-degree
• 𝑓(𝑥) → +∞ as 𝑥 → +∞ and 𝑓(𝑥) → −∞ as 𝑥 → −∞ if 𝑎𝑛 > 0
• 𝑓(𝑥) → −∞ as 𝑥 → +∞ and 𝑓(𝑥) → +∞ as 𝑥 → −∞ if 𝑎𝑛 < 0
• Range is ℝ. (Cannot have an absolute extremum)
•
Must have at least one x-intercept.
Finite Differences
Recall: The first differences of a linear function are constant, as are the second differences of a quadratic
function.
What conjecture might you make about polynomials of higher degree?
Watch Waterloo video on finite differences until 5:40.
Note: The nth-differences of an nth degree polynomial are constant. Moreover, the value of the those n-th
differences when x increases by one, is given by
𝑎𝑛 (𝑛)(𝑛 − 1)(𝑛 − 2) ⋯ (3)(2)(1) OR 𝑎𝑛 𝑛!
[(𝑛)(𝑛 − 1)(𝑛 − 2) ⋯ (3)(2)(1) is denoted 𝑛!]
Ex.: finite 4th differences of 𝑦 = −3𝑥 4 + 3𝑥 3 − 5𝑥 − 7 given by (−3)4! = (−3)(4)(3)(2)(1) = −72
Symmetry – Even and Odd Functions
Reference the previous day’s homework.
Def.: 1. A function, f, is said to be even if 𝑓(−𝑥) = 𝑓(𝑥). An even function is symmetric about the
y-axis. (e.g. 𝑦 = 𝑥 2 is even)
2. A function, f, is said to be odd if 𝑓(−𝑥) = −𝑓(𝑥). An odd function is symmetric on a 180o rotation
about the origin. (e.g. 𝑦 = 𝑥 3 is odd)
Ex.:
1) 𝑦 = −2𝑥 5 + 15𝑥 is an odd function
2) 𝑦 = 2𝑥 4 − 3𝑥 2 − 𝑥 + 2 is neither even nor odd
2
3) 𝑦 = 2𝑥 −1 is an even function
Note: Careful not to confuse even and odd functions with even and odd degreed polynomials.
Ex.:
𝑦 = 𝑥 3 + 1 is an odd-degreed polynomial, but not an odd function
𝑦 = 𝑥 2 + 𝑥 is an even degreed polynomial but not an even function
𝑦 = 𝑥 6 − 𝑥 2 is both an even degreed polynomial and an even function
Transformations of Functions
Note: 1. Vertical Translations: (𝑥) → 𝑓(𝑥) + 𝑘 , (𝑥, 𝑦) → (𝑥, 𝑦 + 𝑘)
• Moves k units upward if 𝑘 > 0
• Moves |𝑘| units down if 𝑘 < 0.
2. Horizontal Translations: 𝑓(𝑥) → 𝑓(𝑥 − ℎ) , (𝑥, 𝑦) → (𝑥 + ℎ, 𝑦)
• Moves h units to the right if ℎ > 0.
• Moves |ℎ| units to the left if ℎ < 0.
3. Reflection in the x-axis: 𝑓(𝑥) → −𝑓(𝑥), (𝑥, 𝑦) → (𝑥, −𝑦)
• Reflected in the x-axis.
4. Reflection in the y-axis: 𝑓(𝑥) → 𝑓(−𝑥), (𝑥, 𝑦) → (−𝑥, 𝑦)
• Reflected in the y-axis.
5. Vertical Stretch: 𝑓(𝑥) → 𝑎𝑓(𝑥), (𝑥, 𝑦) → (𝑥, 𝑎𝑦)
• Expansion by a factor of a if a > 1.
• Compression by a factor of a if 0 < a < 1.
[assume a > 0]
1
6. Horizontal Stretch: 𝑓(𝑥) → 𝑓(𝑏𝑥), (𝑥, 𝑦) → (𝑏 𝑥, 𝑦)
•
Compression by a factor of
•
Expansion by a factor of
1
𝑏
1
𝑏
[assume b > 0]
if b > 1.
if 0 < b < 1.
Note: A general transformation of function, f, called the parent function, is given by
𝑦 = 𝑎𝑓(𝑏(𝑥 − ℎ)) + 𝑘.
•
•
When describing or applying transformations on a parent function, one must consider them in a specific
order: multiplicative ones first (stretches, reflections) and additive ones last (translations).
The input of a transformed function must be in factored form in order to correctly interpret the
3
horizontal transformations [e.g. 𝑦 = (3𝑥 + 9)3 → 𝑦 = (3(𝑥 + 3)) ]
•
A point (𝑥, 𝑦) on f gets transformed to the point (𝑏 𝑥 + ℎ, 𝑎𝑦 + 𝑘).
1
Zeroes of a Polynomial
Def.: The x-intercepts of a function are also referred to as the zeroes of the function.
Note: 1. The zeroes of a function, f, can be determined by solving the equation 𝑓(𝑥) = 0 (the roots).
2. If 𝑥 − 𝑥𝑜 is a factor of a polynomial, then xo is a zero of the polynomial.
b
3. Alternately, if 𝑎𝑥 − 𝑏 is a factor of a polynomial, then
is a zero of the polynomial.
a
Def.: 1. If (𝑥 − 𝑎) is a factor of polynomial, then a is a zero of order 1.
2. If (𝑥 − 𝑎)2 is a factor of polynomial, then a is a zero of order 2.
3. If (𝑥 − 𝑎)3 is a factor of polynomial, then a is a zero of order 3.
…and so on.
Note: Let P be a polynomial.
1. If a is a zero of P of order one, then f crosses the x-axis at a similar to a line.
2. If a is a zero of P of order two, then f touches the x-axis at a similar to a quadratic.
3. If a is a zero of P of order three, then f crosses the x-axis at a similar to a cubic.
4. A similar pattern persists for zeroes of higher orders.
Note:
•
•
•
•
Procedure for sketching a polynomial function
Determine the degree and sign of the leading coefficient to determine the end behaviour of the function
Determine the x-intercepts by factoring or other means (e.g. quadratic formula, other methods to come)
Determine the order of the zeroes
Sketch the graph using the information above (determining other features of the graph will require
calculus tools)
Families of Polynomial Functions
Def.: The set of polynomial function of degree n with the same zeroes, that is
𝑓(𝑥) = 𝑘(𝑥 − 𝑧1 )(𝑥 − 𝑧2 ) ⋯ (𝑥 − 𝑧𝑛 ) , 𝑘 ∈ ℝ\{0}
is called the family of polynomial with zeroes z1 , , zn .
We can also consider the family of polynomials of degree n with zeroes of various order:
𝑓(𝑥) = 𝑘(𝑥 − 𝑧1 )𝑝1 (𝑥 − 𝑧2 )𝑝2 ⋯ (𝑥 − 𝑧𝑚 )𝑝𝑚 , 𝑘 ∈ ℝ\{0} and 𝑝1 + 𝑝2 + ⋯ + 𝑝𝑚 = 𝑛
(Note that functions in a family are related by vertical stretches and/or reflections in the x-axis)
Note: To determine the equation of a polynomial (of least degree) given its zeroes, with their orders, require
you to know another point on the function. To determine the equation:
• Write the family of polynomials based on the zeroes given
• Input the other know point and solve for k
Division of Polynomials
Ex.: Consider the function 𝑦 = 𝑥 3 − 2𝑥 2 − 5𝑥 + 6. Our factoring techniques up to this point do not work.
Recall long division of numbers:
The result of dividing a number (the dividend) by another number (the divisor) can be expressed as
follows:
𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑
𝑑𝑖𝑣𝑖𝑠𝑜𝑟
= 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 +
𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
𝑑𝑖𝑣𝑖𝑠𝑜𝑟
You can check that your answer is correct by re-writing the expression in the following form (by
multiplying both sides by the divisor):
𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑 = (𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡)(𝑑𝑖𝑣𝑖𝑠𝑜𝑟) + 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
If the remainder if 0 then the divisor is a factor of the dividend, and so we can write
𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑
𝑑𝑖𝑣𝑖𝑠𝑜𝑟
= 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡
or
𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑 = (𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡)(𝑑𝑖𝑣𝑖𝑠𝑜𝑟)
Note: 1. Let 𝑃(𝑥) be a polynomial of degree n and 𝐷(𝑥) be a polynomial of degree 𝑚 ≤ 𝑛, then
𝑃(𝑥)
𝐷(𝑥)
= 𝑄(𝑥) +
𝑅(𝑥)
𝐷(𝑥)
OR 𝑃(𝑥) = 𝐷(𝑥)𝑄(𝑥) + 𝑅(𝑥)
where the deg(Q) = n – m and deg(R) < deg(D).
P is called the dividend, D is the divisor, Q is the quotient and R is the remainder.
2. In particular, if D is linear then the remainder will be a constant.
𝑃(𝑥)
𝑥−𝑎
= 𝑄(𝑥) +
𝑅
𝑥−𝑎
OR
𝑃(𝑥) = 𝑄(𝑥) (𝑥 − 𝑎) + 𝑅
Note that if (𝑥 − 𝑎) is a factor of P, then 𝑅 = 0.
Note: Synthetic division can be used to divide a polynomial by a linear polynomial of the form 𝑥 − 𝑎.
It may also be adapted when dividing by a linear expression of the form 𝑎𝑥 − 𝑏, but you must first
𝑏
factor out the a, and then divide by 𝑥 − 𝑎 (see example).
The Remainder Theorem
Ex.: Consider a polynomial P divided ( x − a) , then
𝑃(𝑥) = (𝑥 − 𝑎)𝑄(𝑥) + 𝑅
Notice that 𝑃(𝑎) = (𝑎 − 𝑎)𝑄(𝑎) + 𝑅, ie.
𝑃(𝑎) = 𝑅
Similarly when dividing P by (𝑎𝑥 − 𝑏)
𝑃(𝑥) = (𝑎𝑥 − 𝑏)𝑄(𝑥) + 𝑅
𝑏
𝑏
We see that 𝑃 (𝑎) = (𝑎 (𝑎) − 𝑏) + 𝑅, i.e.
𝑏
𝑃 (𝑎) = 𝑅
Note: The Remainder Theorem
If a polynomial P is divided by (𝑥 − 𝑎), then the remainder R is equal to 𝑃(𝑎).
𝑏
Alternately, if a polynomial P is divided by (𝑎𝑥 − 𝑏), then the remainder R is equal to 𝑃 (𝑎).
The Factor Theorem
Ex.:
Consider the scenario where (𝑥 − 𝑎) is a factor of a polynomial P, that is
𝑃(𝑥) = (𝑥 − 𝑎)𝑄(𝑥), and so
𝑃(𝑎) = (𝑎 − 𝑎)𝑄(𝑎)
𝑃(𝑎) = 0, ie. a is a zero of P.
On the other hand, if 𝑃(𝑎) = 0, then the remainder of the division 𝑃(𝑥) ÷ (𝑥 − 𝑎) is zero. That is
𝑃(𝑥) = (𝑥 − 𝑎)𝑄(𝑥), hence (𝑥 − 𝑎) is a factor of P.
Note: The Factor Theorem
(𝑥 − 𝑎) is a factor of a polynomial P if and only if (iff) 𝑃(𝑎) = 0.
𝑏
Alternately, (𝑎𝑥 − 𝑏) is a factor of a polynomial P if and only if (iff) 𝑃 (𝑎) = 0.
How can we determine zeroes of P in order to apply the Factor Theorem?
Guess at them, but guess intelligently.
Note: If (𝑥 − 𝑎), 𝑎 ∈ ℤ, is a factor of a polynomial P then a must be a factor of P’s constant term.
What if a polynomial has a factor of the form (𝑎𝑥 − 𝑏), 𝑎, 𝑏 ∈ ℤ? How might use the factor theorem in that
circumstance? Consider a polynomial P, what conditions can be placed on the possible values for a and b?
𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 = (𝑎𝑥 − 𝑏)(𝑐𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑐1 𝑥 + 𝑐0 )
We have that 𝑎𝑛 = 𝑎𝑐𝑛−1, and 𝑎0 = −𝑏𝑐0 , ie. a must be a factor of the leading term and b must be a
factor of the constant term.
Note: If (𝑎𝑥 − 𝑏), 𝑎, 𝑏 ∈ ℤ, is a factor of a polynomial P then a must be a factor of the leading coefficient of
P and b must be a factor of the constant coefficient of P.
(Therefore, test inputs of the form
𝑏
𝑎
to determine the zeroes of the polynomial where a is a factor of the
leading coefficient and b is a factor of the constant coefficient.)
Solving Polynomial Equations
Note: To solve a polynomial equation, we can bring all terms to one side of the equation, leaving zero on the
other side, and then factor the side with the polynomial expression. The solution(s) will the zeroes of the
individual factors.
Algorithm:
• Move all terms to one side of the equation
• Factor out common factors
• Factor using traditional methods (e.g. difference of squares, trinomial factoring, grouping, change
of variable)
• Use the Factor Theorem to find linear factors, then divide those out and continue factoring the
quotient (Note that once you arrive at quadratic factors you may also use the Quadratic Formula)
• Use a graphing calculator to estimate the zeroes for polynomials that are not factorable by the above
methods
Solving Polynomial Inequalities
Note: A polynomial function can only change sign at an x-intercept. However, it need not change sign at an xintercept (e.g. zero with and even order).
Note: Algorithm for solving a polynomial inequality:
• Bring all terms to one side and factor
• Use the sign of the leading coefficient, the degree of the function, and the order of the zeroes to sketch
a quick graph to determine when the polynomial will be positive or negative.
• Give your answer in set notation, interval notation or using a number line
• (Be careful when considering the endpoints of an interval: a zero is only included with ≤ and ≥)
Linear Inequalities
Note: 1. An inequality remains true if you:
• Add or subtract a number from both sides.
• Multiply or divide a positive number to both sides.
2. If you multiply or divide an inequality by a negative number, then the inequality will be reversed.
3. A linear inequality can be solved in the same way as a linear equation, by using reverse operations.
(Remember that the inequality must be reversed when multiplying or dividing each side by a negative
number!)
Sum and Difference of Cubes
Note: 1. (𝐴3 − 𝐵 3 ) = (𝐴 − 𝐵)(𝐴2 + 𝐴𝐵 + 𝐵 2 )
2. (𝐴3 + 𝐵 3 ) = (𝐴 + 𝐵)(𝐴2 − 𝐴𝐵 + 𝐵 2 )
Ex:
.
(𝑥 6 − 8) = ((𝑥 2 )3 − (2)3 ) = (𝑥 2 − 2)(𝑥 4 + 2𝑥 2 + 4)
(27𝑥 3 + 1) = ((3𝑥)3 + (1)3 ) = (3𝑥 + 1)(9𝑥 2 − 3𝑥 + 1)
Unit 2: Rational Functions and Equations
Def.: A rational function is a function obtained by the quotient of two polynomial functions. That is, h is a
rational function if ℎ(𝑥) =
𝑓(𝑥)
, where f and g are both polynomials.
𝑔(𝑥)
Reciprocal Functions
Def.: Let f be a function then its reciprocal function refers to the multiplicative inverse of f, that is
𝑔(𝑥) =
1
𝑓(𝑥)
Ex.:
. (If f is a polynomial, then a reciprocal function is also a rational function)
If 𝑓(𝑥) = 𝑥 2 − 7, then the reciprocal function is 𝑔(𝑥) =
If 𝑓(𝑥) = 2𝑥 , then the reciprocal function is 𝑔(𝑥) =
1
2𝑥
1
.
𝑥 2 −7
.
Note: Graphing the reciprocal of a function f:
1. Determine the x-intercepts of f. These will be the locations of the vertical asymptotes.
2. Invariant points occur when f ( x) = 1 . (Invariant points are points that don’t change when a
function undergoes a transformation.)
3. If 𝑓 → +∞, then
4. If 𝑓 → −∞, then
1
𝑓
1
𝑓
will approach 0 from the positive side.
will approach 0 from the negative side.
5. If 𝑓 → 0 from the positive side then
6. If 𝑓 → 0 from the negative side then
7. Reciprocal functions, 𝑦 =
1
𝑓(𝑥)
1
𝑓
1
𝑓
→ +∞.
→ −∞.
, will not have any x-intercepts.
8. Local maxima on f will result in local minima on
1
on .
𝑓
1
𝑓
and local minima on f will result in local maxima
9. The sign of 𝑓(𝑥) and
1
𝑓(𝑥)
are always the same.
Rational Functions
[ In this section, 𝒇(𝒙) and 𝒈(𝒙) will represent polynomials ]
Note 1: In order to analyse a rational function you must first simplify any common factors in the numerator
and denominator.
• If a factor of (𝑎𝑥 − 𝑏) is reduced from the numerator and denominator, then there will be a hole at
b
x = (and so it will not be in the domain).
a
• Remember to keep this restriction throughout your analysis of the function.
[For the rest of this section, assume ℎ(𝑥) =
𝑓(𝑥)
𝑔(𝑥)
Note 2: The zeroes of a rational function ℎ(𝑥) =
Note 3: The function ℎ(𝑥) =
𝑓(𝑥)
𝑔(𝑥)
has already been simplified.]
𝑓(𝑥)
𝑔(𝑥)
are the same as the zeroes of 𝑓(𝑥).
has vertical asymptotes at all x where 𝑔(𝑥) = 0.
[Recall:
• A function f has a vertical asymptote at 𝑥 = 𝑎 if 𝑓(𝑥) → ±∞ as 𝑥 → 𝑎+ (from the right) or as 𝑥 → 𝑎−
(from the left).
• A function f has a horizontal asymptote 𝑦 = 𝑏 if 𝑓(𝑥) → 𝑏 as 𝑥 → −∞ or as 𝑥 → +∞.]
Note 4: If deg(𝑔) > deg(𝑓) then the function ℎ(𝑥) =
𝑓(𝑥)
𝑔(𝑥)
has a horizontal asymptote of 𝑦 = 0. That is, as
𝑥 → ±∞, ℎ(𝑥) → 0.
Note 5: If deg(𝑔) = deg(𝑓), then the function ℎ(𝑥) =
𝑓(𝑥)
𝑔(𝑥)
has a horizontal asymptote of 𝑦 =
𝑎𝑓
𝑎𝑔
, where 𝑎𝑓
is the leading coefficient of f and 𝑎𝑔 is the leading coefficient of g. That is, as → ±∞, ℎ(𝑥) →
Note 6: If deg(𝑔) < deg(𝑓) then the function ℎ(𝑥) =
𝑓(𝑥)
𝑎𝑓
𝑎𝑔
.
, as 𝑥 → ±∞, ℎ(𝑥) → ±∞. (More details to
𝑔(𝑥)
follow!)
Note: Justifying the end behaviour of rational functions:
1. In terms of end behaviour, polynomials of a higher degree have higher growth rate (or rate of
decrease) than polynomial of lower degree (e.g. 𝑦 = 𝑥 3 grows faster than 𝑦 = 𝑥 2 + 100𝑥 in the
long run)
2. The above justifies notes (4) and (6)
3. To justify note (5), where numerator and denominator have similar growth rates Since they are of
equal degrees), you can divide numerator and denominator by the highest power of x. All terms
except the leading terms will tend to 0 as 𝑥 → ±∞
e.g.
𝑦=
3𝑥 3 −2𝑥 2 −7
4𝑥 3 +𝑥
3𝑥3 −2𝑥2 −7
)
𝑥3
4𝑥3 +𝑥
( 3 )
𝑥
2
(
=
=
7
3−𝑥− 3
𝑥
1
4+ 2
𝑥
3
→ 4 as 𝑥 → ±∞
Thus justifying that that the function approached the ratios of the leading coefficients
4. Consider note (6) again. While we have justified that ℎ(𝑥) → ±∞ as 𝑥 → ±∞. We can be more
precise, we can describe how it approaches ±∞:
•
ℎ(𝑥) =
𝑓(𝑥)
𝑔(𝑥)
will approach the quotient of 𝑓(𝑥) ÷ 𝑔(𝑥) as 𝑥 → ±∞ (that is, it will approach
a polynomial function of degree = deg(𝑓) − deg (𝑔)
•
In particular, if deg(f ) = deg(g) + 1, then h will approach a line called a linear oblique
asymptote.
Def.: A function has a discontinuity at 𝑥𝑜 if there is a “break in the continuity” of the function. Examples
include vertical asymptotes, holes and jumps.
Note: A function can only change sign at an x-intercept or at a discontinuity.
𝑥−1
• Ex.: 𝑦 = 2
can only change sign at 𝑥 = 1 (x-intercept) or 𝑥 = ±2 (location of asymptotes)
𝑥 −4
•
That is, if a function f has no discontinuities on an interval (𝑎, 𝑏), then f must have the same sign for all
𝑥 ∈ (𝑎, 𝑏).
•
Note that a function does not have to change sign at a discontinuity.
•
To test the sign of a function over its domain, it suffices to test the sign of one value in each interval
produced by the discontinuities
𝑥−1
o Ex.: To determine the sign of 𝑦 = 2
, test the sign of y for 1 value in each of the intervals
𝑥 −4
(−∞, −2), (−2,1), (1,2), (2, +∞). (Hint: it is usually easier to use the factored form of the
function)
Note: Algorithm for Graphing Rational Functions
1. Factor the numerator and denominator and simplify if possible. Use this expression to identify the
domain of the function.
2. Any linear factors (𝑎𝑥 − 𝑏) that reduce from numerator and denominator will result in a hole at
𝑏
𝑥= .
𝑎
3. From this point forward, analyse the simplified function ℎ(𝑥) =
𝑓(𝑥)
.
𝑔(𝑥)
4. Solve 𝑓(𝑥) = 0 to obtain the x-intercepts.
5. Solve 𝑔(𝑥) = 0 to obtain the location of the vertical asymptotes.
6. Input 𝑥 = 0 to determine the y-intercept.
7. Determine the end behaviour of the function (horizontal asymptotes, oblique asymptotes, etc.).
8. Determine the sign of ℎ(𝑥) in the intervals created by the x-intercepts and discontinuities.
9. Set up the Cartesian plane by choosing a scale, plotting the x and y intercepts, drawing all
asymptotes.
10. Use the results of (8) to sketch the graph of the function (Don’t forget the holes!)
Solving Rational Equations
Note: Algorithm for solving a rational equation:
•
•
•
Multiply both sides of the equation by the lowest common denominator (in order to clear all
denominators)
Solve the resulting polynomial equation
Remember to keep track of restrictions from the original denominators
Solving Rational Inequalities
Note: A rational function can only change sign at an x-intercept, a vertical asymptote or a hole located on the
x-axis. The function may or may not change sign at these locations
Note: Algorithm for solving a rational inequality:
•
•
•
•
•
•
•
Bring all terms to one side
Simplify to one rational expression
Determine the zeroes and locations of the vertical asymptotes (and holes)
Test the sign of the function at one x-value in each intervals produced by the zeroes and the vertical
asymptotes (and holes)
Determine when the inequality is satisfied
Give your answer in set notation, interval notation or using a number line
(Be careful when considering the endpoints of an interval: a zero is only included with ≤ and ≥; the
location of a vertical asymptote will never be included)
Unit 3: Exponential and Logarithmic Functions
Exponents and Exponential Functions Review
1. Properties of exponents:
•
•
𝑎𝑛 = 𝑎 × 𝑎 × ⋯ × 𝑎 (n times), for 𝑛 ∈ ℕ\{0}
𝑎0 = 1, for 𝑎 ∈ ℝ\{0}
•
𝑎 𝑛 = √𝑎
•
𝑎−𝑛 =
1
𝑛
1
𝑎𝑛
•
•
•
•
𝑎 can be defined for all 𝑥 ∈ ℝ
𝑎 𝑥 × 𝑎 𝑦 = 𝑎 𝑥+𝑦
𝑎 𝑥 ÷ 𝑎 𝑦 = 𝑎 𝑥−𝑦
(𝑎 𝑥 )𝑦 = 𝑎 𝑥×𝑦
•
𝑎 𝑛 = √𝑎 𝑚 = ( √𝑎 )
𝑥
𝑚
𝑛
𝑛
𝑚
2. Exponential Functions:
𝑦 = 𝑎 𝑥 , 𝑎 > 0, 𝑎 ≠ 1
• Domain: ℝ
• Passes through (0,1)
• Horizontal asymptote at 𝑦 = 0
• y>0
•
•
•
For 𝑎 > 1:
o y increases over the domain
o End behaviour: as 𝑥 → +∞, 𝑦 → +∞ and as 𝑥 → −∞, 𝑦 → 0
For 0 < 𝑎 < 1:
o y decreases over the domain
o End behaviour: as 𝑥 → +∞, 𝑦 → 0 and as 𝑥 → −∞, 𝑦 → +∞
Exponential functions can be transformed as other functions can: 𝑦 = 𝑘𝑎𝑏(𝑥−𝑑) + 𝑐
3. Applications of Exponential Functions:
•
•
Use to solve exponential growth/decay problems (e.g. half life, compound interest,
appreciation/depreciation)
Common form for growth and decay problems 𝑦 = 𝑘𝑎 𝑥 (k being the initial value)
Inverse Functions Review
•
•
•
•
The inverse of a function 𝑦 = 𝑓(𝑥) is denoted 𝑦 = 𝑓 −1 (𝑥)
o Note that the inverse is NOT the reciprocal (the reciprocal represents the multiplicative inverse,
the notation above represents the functional inverse)
o Note that the inverse is not necessarily a function (e.g. the inverse of 𝑦 = 𝑥 2 is 𝑦 = ±√𝑥, which
does not represent a function. However the domain of f can be restricted so that 𝑓 −1 is a
function (e.g. the inverse of 𝑦 = 𝑥 2 , 𝑥 ∈ [0, +∞) is 𝑦 = √𝑥 which is a function.
Numerical perspective:
o The ordered pairs of 𝑓 −1 are obtained by interchanging the x and y coordinates of the ordered
pairs of f
Graphical perspective:
o The graph of 𝑦 = 𝑓 −1 (𝑥) is a reflection of the graph of 𝑦 = 𝑓(𝑥) about the line y = x .
Algebraic perspective:
o Interchange the x and y variables in the equation of 𝑦 = 𝑓(𝑥) and solve for y to get the equation
for 𝑦 = 𝑓 −1 (𝑥).
Note: Let f be a function, then
𝑓 −1 (𝑓(𝑥)) = 𝑥
and
𝑓(𝑓 −1 (𝑥)) = 𝑥
That is, if a function and its inverse are applied consecutively, in either order, then the output is the
same as the input.
Logarithmic Functions
Def.: The inverse function of the exponential function 𝑦 = 𝑎 𝑥 , 𝑎 > 0, 𝑎 ≠ 1, is called the logarithmic
function with base a and is denoted 𝑦 = log 𝑎 𝑥
We define a logarithmic function by its relationship to the associated exponential function.
𝑦 = 𝑎𝑥
← exponential function
𝑦
𝑥=𝑎
← the inverse of the exponential (exponential form)
𝑦 = log 𝑎 𝑥
← solve for y (using new logarithmic notation) – this is a logarithmic function
Note: 𝑦 = log 𝑎 𝑥 is equivalent to 𝑎 𝑦 = 𝑥. That is, log 𝑎 𝑥 is equal to the power of a that will produce x.
Ex.: 𝑦 = log 3 27 is equivalent to 3𝑦 = 27. And so, log 3 27 = 3
Note: Properties of the logarithmic function 𝑦 = log 𝑎 𝑥:
1. Passes through (1,0).
x
2. Domain = {𝑥 ∈ ℝ|𝑥 > 0} (the range of y = a ).
3. Range = ℝ (the domain of y = a )
4. If a > 1
• 𝑓 → +∞ as 𝑥 → +∞ as x → + .
• 𝑓 → −∞ as 𝑥 → 0+ (that is, a vertical asymptote at x = 0)
• Increases over its domain
5. If 0 < a < 1
• 𝑓 → −∞ as 𝑥 → +∞.
• 𝑓 → +∞ as 𝑥 → 0+ (that is, a vertical asymptote at x = 0)
• Decreases over its domain
x
Note: Since 𝑦 = 𝑎 𝑥 and 𝑦 = log 𝑎 𝑥 are inverses of each other
1. log a (𝑎 𝑥 ) = 𝑥.
2. 𝑎loga 𝑥 = 𝑥
Laws of Logarithms
Note: Laws of Logarithms
1. log 𝑎 (𝑥𝑦) = log 𝑎 𝑥 + log 𝑎 𝑦
𝑥
2. log 𝑎 (𝑦) = log 𝑎 𝑥 − log 𝑎 𝑦
3. log 𝑎 (𝑥 𝑦 ) = 𝑦 log 𝑎 𝑥
[ The “ski-slope rule”
]
Proof of the 1st law:
Let 𝑚 = log 𝑎 𝑥 and 𝑛 = log 𝑎 𝑦
𝑎𝑚 = 𝑥 and 𝑎𝑛 = 𝑦
𝑥𝑦 = 𝑎𝑚 × 𝑎𝑛
𝑥𝑦 = 𝑎𝑚+𝑛
log 𝑎 𝑥𝑦 = log 𝑎 𝑎𝑚+𝑛
log 𝑎 𝑥𝑦 = 𝑚 + 𝑛
log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦
Rewrite in exponential form
Multiply left and right sides
Apply exponent rule to the left side
Take loga of both sides
Simplify right-hand side
Substitute m and n
Proof of the 3rd law:
Let log 𝑎 𝑥 = 𝑚.
𝑎𝑚 = 𝑥
𝑚
(𝑎 )𝑦 = 𝑥 𝑦
𝑎 𝑦𝑚 = 𝑥 𝑦
log 𝑎 𝑎 𝑦𝑚 = log 𝑎 𝑥 𝑦
𝑦𝑚 = log 𝑎 𝑥 𝑦
𝑦 log 𝑎 𝑥 = log 𝑎 𝑥 𝑦
Rewrite in exponential form.
Take both sides to the power of y.
Apply exponent law to the left side.
Take loga of both sides.
Use the fact that log 𝑎 𝑥 and 𝑎 𝑥 are inverses.
Recall that log 𝑎 𝑥 = 𝑚, so
Ex.: Solving an exponential equations using the ski slope rule:
2𝑥 = 5
log 2𝑥 = log 5
Take log10 of both sides
Use ski-slope rule
𝑥 log 2 = log 5
log 5
𝑥=
log 2
Solve for x
This can now be approximated with a calculator
Converting Bases
Note: The logarithm with base 10 is widely used and given its own notation:
log 𝑥 is defined to represent log10 𝑥
Note: Change of Base Formula
log b x
log a x =
.
log b a
log x
In particular, log a x =
.
log a
Proof of change of base formula:
𝑚 = log 𝑎 𝑥
𝑎𝑚 = 𝑥
log 𝑏 𝑎𝑚 = log 𝑏 𝑥
𝑚 log 𝑏 𝑎 = log 𝑏 𝑥
log𝑏 𝑥
𝑚=
log𝑏 𝑎
log𝑏 𝑥
log 𝑎 𝑥 =
log𝑏 𝑎
Let
Rewrite in exponential form.
Take logb of both sides.
Use the ski-slope rule on the left.
Solve for m.
Recall that 𝑚 = log 𝑎 𝑥. So,
Ex.: Solving an exponential equation using a change of base
2𝑥 = 5
Take log 2 of both sides
𝑥 = log 2 5
Use change of base formula to log10
log 5
𝑥=
This can now be approximated with a calculator
log 2
Solving Exponential Equations
Note: Strategies for solving exponential equations
• Write the equation as an equality of two powers with the same base, then equate the exponents
• Write the equation with a power on one, or both, sides of the equation and take a logarithm of both
sides
• Use previous equation solving methods (e.g. factoring, change of variable, etc) to help obtain one of the
above scenarios
• If the above techniques do not apply, use technology
Solving Logarithmic Equations
Note: Strategy for solving logarithmic equations
• Use the laws of logarithms to rewrite the equation with a single logarithm on one side and a constant on
the other
• Once in the form given above, rewrite the equation in exponential form and solve the resulting equation
• Be sure to confirm that the answers found satisfy the expressions in the original equation (e.g. the
inputs in the original expressions must be > 0). Eliminate any that don’t.
•
If the above does not work, use technology
Note: A solution to an equation which does not satisfy an expression in the original equation, and therefore
must not be considered, is called extraneous or an extraneous root.
Exs.: A solution to a logarithmic equation which produced a negative input to a logarithmic
expression
A solution to a rational equation which produces a 0 in the denominator of an expression in the
original equation
Applications of Logarithms
Note: Some characteristics of logarithms which make them useful:
• They transform very large numbers into small, manageable numbers – this is used in logarithmic scales
• They increase at an ever decreasing rate (slower than the other functions we have studied) – they can be
used to model behaviour that grows very slowly
• They turn products into sums, quotients into differences and powers into single products – hence the
use of log and antilog tables (and slide rules) to perform multiplications, divisions and evaluating
powers
Logarithmic Scales
Note: Some common logarithmic scales.
1. Richter scale:
𝑀 = log
𝐼
𝐼𝑜
where M is a magnitude, Io is the intensity of the base earthquake and I is the intensity of the earthquake being
measured.
2. Decibel scale:
𝛽 = 10 log
𝐼
𝐼𝑜
where 𝛽 is the sound level in decibels (dB), Io is the minimum intensity for the human ear ( 10−12 W/m2) and I is
the intensity of the sound being measured. (Sound level is often denoted with L instead of 𝛽)
(The decibel scale can also be defined as follows:
Let 𝛽1 and 𝛽2 be the sound levels, in decibels, of 2 sounds, and 𝐼1 and 𝐼2 be their sound
intensities in W/m2, then
𝐼
𝛽2 −𝛽1 = 10 log 𝐼2 )
1
3. pH scale:
pH = log
1
= − log  H + 
 H + 
where  H +  is the concentration of hydrogen ions in moles/litre.
Unit 4: Trigonometric Expressions and Identities
Radian measure
Note: Dividing the angle of one turn around a circle into 360 units (degrees) is useful as 360 has many whole
factors. Therefore one can divide a circle in many ways without having to work with fractions. However, this
number is arbitrary and has no natural significance.
Def.: One radian (rad) is defined as the angle produced (“subtended”) by an arc with a length equal to the
radius. (The size of a radian is independent of the size of the circle.)
Note: 1. The angle of one revolution around a circle is 2 radians.
2. 2 radians = 360o or  radians = 180o.
3. 1𝑜 =
𝜋
180𝑜
4. 1 rad =
rad
180𝑜
𝜋
Note: Angular velocity is defined as the change in the angle of rotation/revolution, usually measured in
radians, per unit time.
Note: Let  be an angle, in radians, subtended by an arc of length a in a circle of radius r. Then,
a
or
𝑎 = 𝑟𝜃
=
r
(Notice that 𝜃 is given as the ratio of two lengths, hence radians are considered dimensionless. When
describing an angle in radians, we often omit unit symbol rad)
Angles defined on the Cartesian Plane
Recall: An angle can be represented on a Cartesian plane as the angle starting from the positive x axis and
ending at a line which radiates from the origin. This ending line is called the terminal arm.
• An angle is positive if it begins at the positive x-axis and moves counter-clockwise. It is negative if it
moves clockwise.
• Two angles with the same terminal arm are called co-terminal.
𝜋
9𝜋
e.g. 4 and 4 are co-terminal.
•
Two angles are co-terminal if the differ by a factor of 2𝜋. That is,
 and ( + 2𝜋k), k  Z , are co-terminal angles.
•
The related acute angle (RAA) of an angle 𝜃 is the angle between the terminal arm of 𝜃 and the x-axis.
Trigonometric Ratios for any Angle
If we only consider angles in the first quadrant, we can still use a right triangle to define the trigonometric
ratios of the angle (note that the opposite and adjacent sides of the triangle are equivalent to x and y coordinates
of a point on the terminal arm, and the hypotenuse can just be found using Pythagoras). This model fails for
𝜋
angles ≥ 2 .
Note: Trigonometric ratios for any angle on the Cartesian plane:
Let (x,y) be any point on the terminal arm of angle  , and let r (> 0) be the distance from the origin to
(x,y), then
𝑦
𝑥
𝑦
𝑟
𝑟
𝑥
sin 𝜃 = 𝑟
cos 𝜃 = 𝑟
tan 𝜃 = 𝑥
csc 𝜃 = 𝑦
sec 𝜃 = 𝑥
cot 𝜃 = 𝑦
Note: 1. Special angles
𝜋
6
:
𝜋
4
𝜋
:
3
:
2. The “CAST” rule is a mnemonic for remembering the signs of the trigonometric ratios in each of the
four quadrants.
Sine/Cosecant are positive in the 1st and 2nd quadrants, negative in the others
Cosine/Secant are positive in the 1st and 4th quadrants, negative in the others
Tangent/Cotangent are positive in the 1st and 3rd quadrants, negative in the others
Note: Determining trig ratios for special angles ≥
•
𝜋
2
𝜋 2𝜋 3𝜋
(2 ,
3
,
4
,…)
If the terminal arm lies on an axis, use the point with 𝑟 = 1, e.g. for 𝜃 =
3𝜋
2
, use the point (0, −1). If
not, see below.
•
Draw a vertical line for a point on the terminal arm to the x-axis that contains the special related acute
angle
•
Use the appropriate special triangle to determine the trig ratio
•
Use the CAST rule to determine the correct sign or use the signs of the x and y values
Equivalent Trigonometric Expressions
Note: The trigonometric ratios repeat themselves every 2𝜋 radians, that is
1. sin 𝑥 = sin(𝑥 + 2𝜋𝑘), 𝑘 ∈ ℤ.
2. cos 𝑥 = cos(𝑥 + 2𝜋𝑘), 𝑘 ∈ ℤ.
3. tan 𝑥 = tan(𝑥 + 2𝜋𝑘), 𝑘 ∈ ℤ. (This will be revisited later)
4. csc 𝑥 = csc(𝑥 + 2𝜋𝑘), 𝑘 ∈ ℤ.
5. sec 𝑥 = sec(𝑥 + 2𝜋𝑘), 𝑘 ∈ ℤ.
6. cot 𝑥 = cot(𝑥 + 2𝜋𝑘), 𝑘 ∈ ℤ. (This will be revisited later)
Note: Related-angle identities
Consider the related angles: 𝜃, 𝜋 − 𝜃, 𝜋 + 𝜃, 2𝜋 − 𝜃 (𝑜𝑟 − 𝜃)
1.
sin( −  ) = sin 
cos( −  ) = − cos 
tan( −  ) = − tan 
csc(𝜋 − 𝜃) = csc 𝜃
sec(𝜋 − 𝜃) = −sec 𝜃
cot(𝜋 − 𝜃) = −cot 𝜃
2.
sin( +  ) = − sin 
cos( +  ) = − cos 
tan( +  ) = tan 
csc(𝜋 + 𝜃) = −csc 𝜃
sec(𝜋 + 𝜃) = −sec 𝜃
cot(𝜋 + 𝜃) = cot 𝜃
3.
sin(2 −  ) = − sin  or
cos(2 −  ) = cos 
tan(2 −  ) = − tan 
sin(− ) = − sin 
cos(− ) = cos 
tan(− ) = − tan 
csc(2𝜋 − 𝜃) = −csc 𝜃 or csc(−𝜃) = −csc 𝜃
sec(2𝜋 − 𝜃) = sec 𝜃 or sec(−𝜃) = sec 𝜃
cot(2𝜋 − 𝜃) = −cot 𝜃 or cot(−𝜃) = −cot 𝜃
Note: Co-related angle identities or Co-function identities
Consider the co-related angles to 𝜃:
1.
𝜋
2
3𝜋
+𝜃,
2
– 𝜃,
3𝜋
2
+𝜃.
𝜋
cos ( 2 − 𝜃) = sin 𝜃
sec (2 − 𝜃) = csc 𝜃
tan (2 − 𝜃) = cot 𝜃
cot (2 − 𝜃) = tan 𝜃
𝜋
𝜋
𝜋
𝜋
sin (2 + 𝜃) = cos 𝜃
csc (2 + 𝜃) = sec 𝜃
cos ( 2 + 𝜃) = − sin 𝜃
sec (2 + 𝜃) = −csc 𝜃
tan (2 + 𝜃) = − cot 𝜃
cot (2 + 𝜃) = − tan 𝜃
𝜋
𝜋
3𝜋
𝜋
𝜋
3𝜋
sin ( 2 − 𝜃) = − cos 𝜃
csc ( 2 − 𝜃) = −sec 𝜃
cos ( 2 − 𝜃) = − sin 𝜃
sec ( 2 − 𝜃) = −csc 𝜃
tan ( 2 − 𝜃) = cot 𝜃
cot ( 2 − 𝜃) = tan 𝜃
3𝜋
3𝜋
4.
𝜋
csc (2 − 𝜃) = sec 𝜃
𝜋
3.
2
– 𝜃,
sin (2 − 𝜃) = cos 𝜃
𝜋
2.
𝜋
3𝜋
3𝜋
3𝜋
3𝜋
sin ( 2 + 𝜃) = − cos 𝜃
csc ( 2 + 𝜃) = −sec 𝜃
cos (
sec (
3𝜋
2
3𝜋
+ 𝜃) = sin 𝜃
tan ( 2 + 𝜃) = − cot 𝜃
3𝜋
2
3𝜋
+ 𝜃) = csc 𝜃
cot ( 2 + 𝜃) = − tan 𝜃
Compound and Double Angle Formulas
Note: Compound Angle Formulas
1. sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
2. sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − sin 𝐵 cos 𝐴
3. cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
4. cos(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
5. tan(𝐴 + 𝐵) =
6. tan(𝐴 − 𝐵) =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
tan 𝐴 − tan 𝐵
1 + tan 𝐴 tan 𝐵
Note: Double Angle Formulas
1. sin(2𝐴) = 2 sin 𝐴 cos 𝐴
2. cos(2𝐴) = cos 2 𝐴 − sin2 𝐴
= 2 cos2 𝐴 − 1
= 1 − 2 sin2 𝐴
3. tan(2𝐴) =
2 tan 𝐴
1−tan2 𝐴
Proving Trigonometric Identities
Def.: An identity is an equality with is true for all values of the variable(s) for which all expressions are
defined.
Note: Recall identities that were established in the previous course:
1. tan 𝜃 =
2. csc 𝜃 =
3. sec 𝜃 =
4. cot 𝜃 =
sin 𝜃
cos 𝜃
1
sin 𝜃
1
cos 𝜃
1
tan 𝜃
=
cos 𝜃
sin 𝜃
5. sin2 𝜃 + cos2 𝜃 = 1
6. tan2 𝜃 + 1 = sec 2 𝜃
7. cot 2 𝜃 + 1 = csc 2 𝜃
Note: Proving identities
1. To prove an identity, you must show that both sides of the equality are equivalent for all values for
which the expressions are defined.
• You may only use previously proven identities to help manipulate expressions (do not make
anything up along the way!).
• Start with one side (it is usually suggested to begin with the more complex side) and manipulate
it until it is equivalent to the other side.
•
•
•
It may be necessary to manipulate both sides of the equality (separately!)
While proving an identity, do not treat the identity as an equation. That is, do not apply
operations to both sides of the equality. To avoid this, never write both sides of the equality on
the same line.
If you cannot prove an identity algebraically, you may also graph both sides as functions and
show that their graphs are equivalent (this is not a proof as you can never illustrate all values of
a function on a graph).
2. To disprove an identity, you need only show that the equality does not hold for one value for which
the expressions are defined (this is called a counterexample).
Unit 5 – Trigonometric Functions and Equations
Defining Trigonometric Functions
Def.: A function is periodic if the y values repeat at regular intervals. If we denote the period of repetition
(the length of one cycle), p, then we can say a function is periodic if
𝑓(𝑥 + 𝑝) = 𝑓(𝑥)
or
𝑓(𝑥 + 𝑛𝑝) = 𝑓(𝑥), 𝑛 ∈ ℤ.
Def.: The functions (𝑥) = sin 𝑥, 𝑓(𝑥) = cos 𝑥, 𝑓(𝑥) = tan 𝑥, 𝑓(𝑥) = csc 𝑥, 𝑓(𝑥) = sec 𝑥 and 𝑓(𝑥) = cot 𝑥
are defined using the angles on the Cartesian plane as the domain values and the trigonometric ratios as
the range values.
Once the functions are defined, the domain and range values can represent any quantities which display
this type of periodic behaviour.
Sinusoidal Functions
Note: 1. The functions 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = cos 𝑥 are referred to as sinusoidal functions. Their graphs
both have wave-like shapes.
2. 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = cos 𝑥 have an absolute maximum value of 1 and an absolute minimum
value of −1.
3. The axis (or sinusoidal axis) of a sinusoidal function is the horizontal line midway between the
maximum and minimum values. That is, the axis is given by
𝑦=
𝑚𝑎𝑥+𝑚𝑖𝑛
2
.
4. The amplitude, a, of a sinusoidal function is the distance between the axis and the maximum or
minimum, or half the distance between the maximum and minimum. That is,
amplitude =
𝑚𝑎𝑥−𝑚𝑖𝑛
2
.
5. The period of 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = cos 𝑥 is 2𝜋.
6. The functions 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = cos 𝑥 have a domain of ℝ and a range of [−1,1].
Non-Sinusoidal Trigonometric Functions
𝜋
Note: 1. 𝑓(𝑥) = tan 𝑥 has a period of 𝜋, a domain of ℝ \ {2 + 𝜋𝑘|𝑘 ∈ ℤ} and a range of ℝ
2. 𝑓(𝑥) = csc 𝑥 has a period of 2𝜋, a domain of ℝ \ {𝜋𝑘|𝑘 ∈ ℤ} and a range of (−∞, 1] ∪ [1, ∞)
𝜋
3. 𝑓(𝑥) = sec 𝑥 has a period of 2𝜋, a domain of ℝ \ { + 𝜋𝑘|𝑘 ∈ ℤ} and a range of (−∞, 1] ∪ [1, ∞)
2
4. 𝑓(𝑥) = cot 𝑥 has a period of 𝜋, a domain of ℝ \ {𝜋𝑘|𝑘 ∈ ℤ} and a range of ℝ
𝑦 = csc 𝑥
Period: 2𝜋
Asymptotes:
𝑥 = 𝜋𝑛, 𝑛 ∈ ℤ
Local mins:
𝜋
( 2 + 2𝜋𝑛, 1) , 𝑛 ∈ ℤ
Local maxs:
𝜋
(− 2 + 2𝜋𝑛, −1) , 𝑛 ∈ ℤ
𝑦 = sec 𝑥
Period: 2𝜋
Asymptotes:
𝜋
𝑥 = 2 + 𝜋𝑛, 𝑛 ∈ ℤ
Local mins:
(2𝜋𝑛, 1), 𝑛 ∈ ℤ
Local maxs:
(𝜋 + 2𝜋𝑛, −1), 𝑛 ∈ ℤ
𝑦 = tan 𝑥
Period: 𝜋
Asymptotes:
𝜋
𝑥 = 2 + 𝜋𝑛, 𝑛 ∈ ℤ
Zeroes:
𝜋𝑛, 𝑛 ∈ ℤ
𝑦 = cot 𝑥
Period: 𝜋
Asymptotes:
𝑥 = 𝜋𝑛, 𝑛 ∈ ℤ
Zeroes:
𝜋
2
+ 𝜋𝑛, 𝑛 ∈ ℤ
Transformations of Sinusoidal Functions
Note: Consider the sinusoidal functions 𝑦 = 𝑎 sin[𝑘(𝑥 − 𝑑)] + 𝑐 and
•
Amplitude = |𝑎|
•
Period = |𝑘| (and so |𝑘|= 𝑝𝑒𝑟𝑖𝑜𝑑 )
•
𝑦 = 𝑐 is the sinusoidal axis
•
The horizontal translation, d, is called the phase shift.
•
Maximum = 𝑐 + |𝑎|
•
Minimum = 𝑐 − |𝑎|
•
|𝑎| =
2𝜋
𝑦 = 𝑎 cos[𝑘(𝑥 − 𝑑)] + 𝑐, then
2𝜋
𝑚𝑎𝑥−𝑚𝑖𝑛
2
𝑚𝑎𝑥+𝑚𝑖𝑛
•
𝑐=
•
If 𝑎 < 0, the function will have undergone a reflection in the x-axis.
•
If 𝑘 < 0, the function will have undergone a reflection in the y-axis.
2
Sketching Trigonometric Functions
Note: Sketching trigonometric functions.
•
Sinusoidal functions:
o Use the parameters a, k, c and d to determine the maximum, minimum, sinusoidal axis, period
and phase shift
o Set up the Cartesian plane: draw horizontal lines for the max, min and axis, choose a scale for
the x-axis by dividing the period into parts
o Sketch the graph without the phase shift by first plotting the maximum and minimum points as
well as those on the axis
o Translate the basic points above by the phase shift, then graph
•
Tangent and cotangent:
𝜋
𝜋
o Transform 3 points [e.g. (− 4 , −1) , (0,0), ( 4 , 1) for tangent] and the vertical asymptotes (e.g.
𝜋
𝑥 = −2 ,𝑥 =
•
𝜋
2
for tangent] from the parent function and then sketch.
Cosecant and secant
o You can transform points and asymptotes as described above
o Alternately, you can use the characteristics of the sine/cosine function with the same
characteristics:
▪ The maxima and minima of sine/cosine will become the minima and maxima of
cosecant/secant
▪ The x values where the sine/cosine function crosses the sinusoidal axis will be the
location of the vertical asymptotes of cosecant/secant
Inverse Trigonometric Functions
Note: 1. The function 𝑦 = sin−1 𝑥, or 𝑦 = arcsin 𝑥, is the inverse function of 𝑦 = sin 𝑥. (𝑦 = sin−1 𝑥 is
equivalent to 𝑥 = sin 𝑦, that is, sin−1 𝑥 is the angle that produces a sine value of x)
𝜋 𝜋
• The domain is [−1,1] and the range is [− 2 , 2 ].
•
That is, the output is an angle in the 1st quadrant if sine is positive and a negative angle in
the 4th quadrant if sine is negative.
•
𝜋 − 𝑥 will give a second solution (since sin(𝜋 − 𝑥 ) = sin 𝑥).
•
If x is a solution, then 𝑥 + 2𝜋𝑛, 𝑛 ∈ ℤ and (𝜋 − 𝑥) + 2𝜋𝑚, 𝑚 ∈ ℤ, will also be solutions.
2. The function 𝑦 = cos −1 𝑥, or 𝑦 = arccos 𝑥, is the inverse function of 𝑦 = cos 𝑥. (𝑦 = cos −1 𝑥 is
equivalent to 𝑥 = cos 𝑦, that is, cos −1 𝑥 is the angle that produces a cosine value of x)
•
The domain is [−1,1] and the range is 0,   .
•
That is, the output is an angle in the 1st quadrant if cosine is positive and an angle in the 2nd
quadrant if cosine is negative.
•
2𝜋 − 𝑥 will give a second solution (since cos(2𝜋 − 𝑥) = cos 𝑥).
•
If x is a solution, then 𝑥 + 2𝜋𝑛, 𝑛 ∈ ℤ and (2𝜋 − 𝑥) + 2𝜋𝑚, 𝑚 ∈ ℤ, will also be solutions.
3. The function 𝑦 = tan−1 𝑥, or 𝑦 = arctan 𝑥, is inverse function of 𝑦 = tan 𝑥. (𝑦 = tan−1 𝑥 is
equivalent to 𝑥 = tan 𝑦, that is, tan−1 𝑥 is the angle that produces a tangent value of x)
𝜋 𝜋
•
The domain is ℝ and the range is (− 2 , 2 ).
•
That is, the output is an angle in the 1st quadrant if tangent is positive and a negative angle in
the 4th quadrant if tangent is negative.
•
𝜋 + 𝑥 will give a second solution (since tan(𝜋 + 𝑥) = tan 𝑥)
•
Recall that tan 𝑥 has a period of 𝜋, hence only one statement is needed to give all possible
values: If x is a solution, then 𝑥 + 𝜋𝑛, 𝑛 ∈ ℤ will also be solutions.
Solving Trigonometric Equations
Note: Algorithm for solving trigonometric equations
•
Isolate the trigonometric ratios in the given equation
o For a linear trigonometric equation, isolate using opposite operations
o For more complex trigonometric equations, you may need to use other methods such as
factoring, change of variable and trigonometric identities
•
Use the inverse trigonometric function or special angles to determine a first answer. Use the
appropriate related angle identity to determine a second answer.
•
Solve for the variable if needed
•
Use the period of the trigonometric expression(s) in the equation to determine further answers
as needed based on the given domain (e.g. for a sine/cosine expression, if 𝑘 = 𝜋, then per. = 2,
keep adding/subtracting 2 to the initial two answers until you go beyond the given domain)
•
The following may useful when trigonometric expressions cannot be isolated:
𝜋
o sin 𝑥 = cos 𝑥 for 𝑥 = 4 + 𝜋𝑛, 𝑛 ∈ ℤ
o sin 𝑥 = − cos 𝑥 for 𝑥 =
•
3𝜋
4
+ 𝜋𝑛, 𝑛 ∈ ℤ
If algebraic methods do not work, bring all terms to one side of the equation, graph it on a
graphing calculator and identify the zeroes as your answers
Unit 6 – Combinations of Functions and Rate of Change
There are many ways of combining functions. We’ve already seen the concept of piecewise functions where
different functions are used over different parts of the domain. Other options are arithmetic combinations of
functions, that is, the basic arithmetic operations of addition, subtraction, multiplication and division.
Arithmetic Combinations of Functions
Def.: Let 𝑓 and 𝑔 be functions, then we define the functions 𝑓 + 𝑔, 𝑓 − 𝑔, 𝑓𝑔 and
𝑓
𝑔
as follows:
(𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥)
(𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥)
(𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥)
𝑓
(𝑔) (𝑥) =
𝑓(𝑥)
𝑔(𝑥)
, where 𝑔(𝑥) ≠ 0
Def.: The intersection of two sets is the set of elements that is common to both sets. The symbol for
intersection is ∩.
Ex.: Let 𝐴 = (−∞, 2) and 𝐵 = [−3,4), then 𝐴 ∩ 𝐵 = [−3,2)
Note: Some characteristics of arithmetic combinations of functions. Let f and g be functions.
1. 𝐷𝑓+𝑔 , 𝐷𝑓−𝑔 , 𝐷𝑓𝑔 = 𝐷𝑓 ∩ 𝐷𝑔
2. 𝐷𝑓 = 𝐷𝑓 ∩ 𝐷𝑔 \ {𝑥 ∈ 𝑹|𝑔(𝑥) ≠ 0}
𝑔
3. (𝑓 + 𝑔)(𝑥) = 0 if 𝑓(𝑥) = −𝑔(𝑥)
4. (𝑓 − 𝑔)(𝑥) = 0 if 𝑓(𝑥) = 𝑔(𝑥)
5. (𝑓𝑔)(𝑥) = 0 if 𝑓(𝑥) = 0 or 𝑔(𝑥) = 0
𝑓
6. (𝑔) (𝑥) = 0 if 𝑓(𝑥) = 0 and 𝑔(𝑥) ≠ 0
𝑓
7. If 𝑔(𝑥) = 0 and 𝑓(𝑥) ≠ 0 then (𝑔) (𝑥) will have an vertical asymptote of 𝑥 = 𝑥𝑜
𝑓
8. If 𝑓(𝑥𝑜 ) = 𝑔(𝑥𝑜 ) = 0, then (𝑔) (𝑥) may have a hole or vertical asymptote at 𝑥 = 𝑥𝑜 , further
investigation is required.
9. The end behaviour of an arithmetic combination of functions must be considered on a case-by-case
basis by considering the end behaviours of the individual functions (e.g. horizontal asymptotes, growth
and decay rates). The behaviour as 𝑥 → −∞ and as 𝑥 → +∞ must often be considered separately.
Recall:
Def.: 1. A function f is said to be even if 𝑓(−𝑥) = 𝑓(𝑥) ∀𝑥 ∈ 𝐷𝑓
2. A function f is said to be odd if 𝑓(−𝑥) = −𝑓(𝑥) ∀𝑥 ∈ 𝐷𝑓
Composition of Functions
Def.: Let f and g be functions. The composition of the functions f and g, denoted 𝑓 ∘ 𝑔, is a new function
defined by:
(𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
•
•
•
That is, for a given x value, the output of g becomes the input of f. (Apply the second function – the one
“closest” to the input – first.)
𝑓 ∘ 𝑔 can also be said “f composed with g” , “f of g” , “f circle g”
We can represent the operation in the following diagram:
(When we considered the operations 𝑓 −1 (𝑓(𝑥)) and 𝑓(𝑓 −1 (𝑥)) with a function and its inverse, we were
performing a composition. These can be written 𝑓 −1 ∘ 𝑓 and 𝑓 ∘ 𝑓 −1 respectively.)
Note: 1. Composition is not commutative, that is, in general, 𝑓 ∘ 𝑔 ≠ 𝑔 ∘ 𝑓.
2. The domain of 𝑓 ∘ 𝑔 is the set of all x values in the domain of g such that 𝑔(𝑥) is in the domain of f.
Using set notation: 𝐷𝑓∘𝑔 = {𝑥 ∈ 𝐷𝑔 |𝑔(𝑥) ∈ 𝐷𝑓 }.
Rate of Change
Def.: The rate of change (ROC) of a function indicates how the dependent variable – y – changes with respect
to a change in the independent variable – x.
Def.: A secant is a line which joins two points on a function.
Def.: The tangent to a curve, 𝑓, at 𝑎 is a line which contains the point (𝑎, 𝑓(𝑎)), and approximates the
curve near that point.
Average Rate of Change
Def.: The average rate of change measures the rate at which the function changes over a given interval.
Def.: The average rate of change of a function on the interval (𝑥1 , 𝑥2 ) is the slope of the secant joining
(𝑥1 , 𝑓(𝑥1 )) and (𝑥2 , 𝑓(𝑥2 )). That is,
𝑓(𝑥2 )−𝑓(𝑥1 )
Ave. rate of change =
𝑥2 −𝑥1
Instantaneous Rate of Change
Def.: The instantaneous rate of change (or just rate of change) refers to the rate of change at a given point,
say at 𝑥 = 𝑎.
Def.: The rate of change of a curve at 𝑎 is the slope of the tangent to the curve at (𝑎, 𝑓(𝑎)).
Note: Ways to estimate the rate of change of a function f at 𝑥 = 𝑎:
• Choose a nearby point to the left or right of 𝑎 and use the slope of the secant joining that point and
(𝑎, 𝑓(𝑎)). Unless otherwise stated, use 𝑎 + 0.01 or 𝑎 − 0.01 to determine the nearby point.
• You could also average out the slopes of both secants described above
• You may use points on either side of 𝑎 use the slope of the secant joining them as your estimate.
• You can make successive approximations using x values that get closer and closer to 𝑎 from both the
left and right. (This would yield the most accurate approximation but would be the most laborious.)
Note: For a given 𝑥 = 𝑎, the slopes of the secants
𝑓(𝑥)−𝑓(𝑎)
x approaches 𝑎 (from both the left and right).
𝑥−𝑎
, approach the slope of the tangent to f at 𝑎, as
An alternate, and generally more common, formulation is: For a given 𝑥 = 𝑎, the slopes of the secants
𝑓(𝑎+ℎ)−𝑓(𝑎)
, approach the slope of the tangent to f at 𝑎, as h approaches 0 (from both the left and
ℎ
right).