Ms H.L. Tarr
1 Precalculus Review
1.1 The Real Number System
The real number system consists of rational numbers and irrational numbers.
Rational numbers are numbers that can be expressed as the ratio, or quotient, of two integers
with the divisor being a non-zero integer. Thus, a rational number is a number which can
be expressed in the form a/b where a and b are integers and b 6= 0. Rational numbers are
characterised by repeating decimals i.e. at some point the decimal, a fixed block of numbers
begins to repeat indefinitely.
137
1 2 23
, ,
,−
5 7 455
750
Because any integer a can be written in the form of a quotient a/1, all integers are also rational
numbers. Zero is also considered to be an integer (neither negative nor positive), and it can be
written in the quotient form 0/b = 0, b 6= 0.
Examples of rational numbers:
Irrational numbers are real numbers which cannot be expressed as the ratio of two integers.
Numbers such
(which is the
√
√ ratio of the circumference of a circle to its
√ as π = 3.14159265...
diameter), 2 = 1.4142..., 3 = 1.7321... and 5 = 2.2361... are all examples of irrational
numbers. Irrational numbers have non-repeating decimals.
In the second semester, we will learn about complex numbers, of which the set of real numbers
is a subset.
The Real Number System
The set of real numbers can be represented using a real number line. The real number line has
a zero point, often called the origin, which is used to represent the real number 0. To each and
every point on the real number line, there corresponds a real number. The correspondence is
that the real number represented by a point equals the directed distance traveled in moving
from the origin to that point. Movements from left to right along the number line are considered
to be in a positive direction. Thus, points to the right of the origin correspond to positive real
numbers, whereas points to the left correspond to negative real numbers. For each and every
real number, there corresponds a unique point on the number line.
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Ms H.L. Tarr
The Real Number Line
The inequality symbols < and > are used to indicate that two numbers are not equal but they
can be compared. When an inequality symbol is placed between two numbers, it ”opens” in the
direction of the larger number (i.e. the sharp point faces towards the smaller number). Given
two real numbers a and b, the notation a > b is read ”a is greater than b”. The statement
implies that, on the real number line, a is located to the right of b.
1.2 Basic Set Theory
Definition 1.1. A set is a collection of objects, called elements of the set.
e.g (a) The set of all members of the Kaiser Chiefs soccer team
(b) The set of all letters of the alphabet
(c) The set of all solutions of a system of equations
(d) The set of all numbers 2, 4, 6, 8, 10, 12
Notation
Capital letters denote sets, and braces (curly brackets {}) group the elements
of a set.
e.g. in (d) above, S = {2, 4, 6, 8, 10, 12}, simply enumerating (writing out
in full) the elements of the set. This is also called ”list notation”.
e.g. in (b) above, S = {A, B, C, ..., Z} (”...” means ”and so on.”)
Small letters a,b,c, denote the elements of a set.
a ∈ S means ”a is an element of the set S” (often read ”a belongs to S”).
a∈
/ S means ”a is not an element of the set S” (or ”a does not belong to S”).
e.g. in (d) above, 2 ∈ S but 3 ∈
/ S.
1.2.1 Special Sets
(1) N = {1, 2, 3, 4, ...} is the set of natural numbers.
(2) Z = {0, ±1, ±2, ±3, ...} is the set of integers (all whole numbers, positive, negative and
zero).
(3) Q denotes the set of all rational numbers i.e. fractions.
(4) R denotes the set of real numbers, consisting of all of Q and all of the irrational numbers
which cannot be written as fractions.
(5) ∅ denotes the empty set or null set which has no elements.
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Ms H.L. Tarr
1.2.2 Descriptive Property Notation (or Set Builder Notation)
Instead of enumeration, we often use the descriptive property notation to describe a set.
e.g. in (d) above, S = {x ∈ R | x = 2n, n ∈ N, n ≤ 6}
Which is read ”S is equal to the set of all x belonging to the real numbers such that x equals
2n and n is an element of the natural numbers and n is less than or equal to 6”.
Notation
(a) | (or :) denotes ”such that” which means ”which satisfy the properties”.
(b) , denotes ”and”.
(c) This notation often involves finding a pattern e.g. in the above example,
noticing that all the elements are positive multiples of 2.
(d) We could have replaced R by Q or Z or N in this example, or left it out
completely.
Exercise 1 If S = { x ∈ R | x > 0 and x < −1 }, then S =
Exercise 2 If S = {x ∈ Z | x ≥ −2 , x < 2}, then S =
4 5
Exercise 3 Express { 32 , 38 , 15
, 24 , ...} in descriptive property form.
1.2.3 Subsets, Intersections and Unions
Definition 1.2. Set A is a subset of set B if every element of A is an element of B. We write
A ⊂ B. If A is not a subset of B we write A 6⊂ B.
If A ⊂ B and B ⊂ A, we say that A equals B, written A = B i.e the set have identical elements.
Clearly, A and ∅ are subsets of A - they are called ”trivial subsets”.
Also, N ⊂ Z ⊂ Q ⊂ R.
A Venn Diagram clearly shows the
relation between A and B. i.e. A ⊂ B
Example 1: If T = {x ∈ Z | 0 ≤ x ≤ 6}
S = {x ∈ N | x2 ≤ 9}
V = {x ∈ Z | x = 3y, y ∈ N, y < 5}
Then S ⊂ T and V 6⊂ T .
Definition 1.3. A ∩ B = {x | x ∈ A and x ∈ B} is called the intersection of sets A and B.
i.e. the set of all elements common to both
set A and set B.
Intersection
e.g. In Example 1 above, S ∩ V = {3} and V ∩ T = {3, 6}
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Ms H.L. Tarr
Definition 1.4. A ∪ B = {x | x ∈ A or x ∈ B or x ∈ A and B} is called the union of sets A
and B.
i.e. the set of all elements of both set A
and set B.
Union
e.g. In Example 1 above, S ∪ V = {1, 2, 3, 6, 9, 12} and V ∪ T = {0, 1, 2, 3, 4, 5, 6, 9, 12}
In general, A ∩ B ⊂ A ⊂ A ∪ B, and similarly for B.
Exercise 4 If A ⊂ B, what can we say about A ∩ B and A ∪ B?
Tutorial 1: Precalculus Review - Set Theory
(1) Find
(b) {x ∈ R | x2 + x − 1 = 0} (c) {x ∈ Q | x2 + x − 1 = 0}
(a) {x ∈ R | x2 = x}
2
(d) {x ∈ R | x + 1 > 0} (e) {x ∈ R | x2 + 1 < 1}
(f) {x ∈ N | x + 1 is divisible by 2}
2x − 3
x
*(h) {x ∈ R | x4 − 2x2 − 2 = 0}
=
(g)
x
x+1
2x − 1
(2) Write in descriptive property form:
1 2 3
, , , ...
(a) S = {5, 10, 15, 20, ..., 1000}
(b) T = {−6, −3, 0, 3, 6} (c) W =
2 3 4
(d) The set of all positive odd numbers (e) Q
(f) V = {1, 5, 9, 13, ...}. Is 41281 ∈ V ?
(3) Let A = {x ∈ R | x2 = 4}, B = {x | x2 − 4x + 4 = 0}, C = {−2,
p2},
D = {x ∈ R | x ≥ 2 and x ≤ 2}, E = {−2, 0, 2}, F = {x | x − 2 − x2 = 1},
2x2 + 3x − 2
x+2
x
and H = x x + 1 =
=
G= x∈R
x−1
x
x+2
Are any of these sets equal ?
What other relationships are there between these sets ?
(4) If A ⊂ B, what can we say about A ∪ B and A ∩ B ? Evaluate N ∪ Z and N ∩ Z.
(5) Express the following sets in descriptive property notation, as simply as possible (Hint:
Find the
pattern!)
2 3 4 5 6 7
100
4 9 16 25 36
(a)
, , , , , , ...
(b)
, , , , , ...,
99
3 8 15 24 35
5 6 7 8 9 10
1 3 5 7 9
19
1 1 1 1 1 1
(d)
, , , , ,
, , , , ,··· ,
(c)
2 5 10 17 26 37
3 5 7 9 11
20
(e) {−5, 10, −15, 20, −25, 30, ..., 50}
Hint: what number, raised to
(f) {5, −10, 15, −20, 25, −30, ..., −50} different powers gives 1 or -1 ?
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Ms H.L. Tarr
Tutorial 1 Solutions
√
√ −1 − 5 −1 + 5
(c) ∅ (d) R
(1) (a) {0, 1} (b)
,
2
2
p
√ p
√
(e) ∅
(f) {1, 3, 5, 7, ...}
(g) ∅ (h) {− 1 + 3}, 1 + 3}
(2) (a) S = {x ∈ N | x = 5k, k ∈ N, k ≤ 200}
(b) T = {x ∈ Z | x = 3y, y ∈ Z, −2 ≤ y ≤ 2} = {x | x = ±3y, y ∈ Z, 0 ≤ y ≤ 2}
n
(c) W = {x ∈ Q | x = n+1
, n ∈ N}
(d) S = {x | x = 2n − 1, n ∈ N} = {x ∈ Z | x = 2t + 1, t ∈ Z, t ≥ 0}
(e) Q = {x ∈ R | x = pq , p, q ∈ Z, q 6= 0}, why must we have q 6= 0 ?
(f) V = {x ∈ N | x = 4n − 3, n ∈ N}, yes - why ?
(3) B = D = F = G = H ⊂ A = C ⊂ E
(4) A ∪ B = B, A ∩ B = A (use a Venn Diagram to see this)
N ∪ Z = Z, N ∩ Z = N
n+1
(5) (a) x ∈ Q | x =
,n ∈ N
n+4
2
(b) {x ∈ Q | x = n2n−1 , n ∈ Z, 2 ≤ n ≤ 10}
(c) {x ∈ Q | x = n21+1 , n ∈ N, n ≤ 6}
(d) {x ∈ Q | x = 2n−1
, n ∈ N, n ≤ 10}
2n+1
(e) {x ∈ Z | x = (−1)n · 5n, n ∈ N, n ≤ 10}
(f) {x ∈ Z | x = (−1)n+1 · 5n, n ∈ N, n ≤ 10}
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Ms H.L. Tarr
1.3 Intervals of the Real Line
Definition 1.5. If a, b ∈ R, with a < b, we define the closed interval
[a, b] = {x ∈ R | a ≤ x ≤ b}.
i.e. the endpoints a and b are included.
Definition 1.6. If a, b ∈ R, with a < b, we define the open interval (a, b) = {x ∈ R | a < x <
b}
i.e. the endpoints a and b are excluded.
Exercise 5 Define [a, b) and (a, b]
N.B In all these intervals, we must have a < b.
We also define (−∞, a) = {x ∈ R | x < a}
And [b, ∞) = {x ∈ R | b ≤ x}
Note: We always have open brackets with −∞ and ∞.
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Ms H.L. Tarr
1.3.1 Combining Intervals
(1) Intersections (”and”) - common points
Example 1: −1 < x and x < 2
∴ −1 < x < 2
i.e. x ∈ (−1, 2)
note endpoints not included
Example 2: −1 ≤ x and x < −3
∴ x ∈ ∅ i.e. no common points
(2) Unions (”or”) - either condition satisfied
Example 3: x > 2 or x ≥ −1
∴ x ≤ −1
i.e. x ∈ [−1, ∞)
Exercise 6 Determine which intervals satisfy the following conditions, write them in descriptive property notation and sketch them.
(a) x ≤ − 12 , x < 0
(b) x > 0 or −2 ≤ x ≤ 0
(c) x < 0 or −2 ≤ x ≤ 0
(d) x ≥ 2 or x < −1
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Ms H.L. Tarr
1.4 Quadratics
1.4.1 Solutions of quadratic equations
An equation of the form y = ax2 + bx + c (a 6= 0) is called a quadratic equation in x. The
graph of this equation is a parabola, whose turning point (T.P.) has an x-coordinate given by
−b
the formula: x =
2a
Definition 1.7. The solutions or roots of an equation are those values of x which satisfy the
equation i.e. those values of x for which the equation holds.
General Method for solving quadratic equations
(1) Put the given equation into the standard form ax2 + bx + c = 0 where a, b, c ∈ Z by:
(a) eliminating fractions
(b) taking all terms to the LHS (left-hand side).
(2) Factorise, if possible.
(3) If not possible to factorise, use the quadratic formula
√
−b ± b2 − 4ac
x=
2a
Solving the quadratic equation in this manner finds the x-intercepts of the parabola, which can
then be sketched.
Example 1: Solve x2 = 3x................(1)
(N.B. Don’t be tempted to divide through by x in equation (1), otherwise you will lose the
solution x = 0.)
Definition 1.8. ∆ = b2 − 4ac is called the discriminant.
Theorem 1.9.
(1) ∆ ≥ 0 ⇒real roots (i.e. roots ∈ R)
• ∆ = 0 ⇒ 2 equal roots
⇒ roots are rational (i.e. roots ∈ Q)
• ∆ a perfect square
(2) ∆ < 0 ⇒ no real roots
Note: We can write a quadratic ax2 + bx + c in the form:
(i) a(x − α1 )(x − α2 ) if it has two distinct real roots
(ii) a(x − α)2
if it has equal roots α.
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Ms H.L. Tarr
1.4.2 Quick graphs of parabolas y = ax2 + bx + c (a 6= 0)
(i) Look at a:
a>0⇒
a<0⇒
(ii) Look at the root (3 Cases)
(a) no real root (∆ < 0) ⇒ graph does not cut the x-axis i.e. no x-intercepts
a>0
a<0
(b) equal roots α(∆ = 0) ⇒ graph touches the x-axis at x = α
a>0
a<0
(c) two distinct roots α1 and α2 (∆ > 0) ⇒ graph cuts the x-axis at α1 and α2
a>0
a<0
Note: Practise this until you can do it quickly -it is useful for solving inequalities!
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Ms H.L. Tarr
Exercise 7 Sketch the graph of the following parabolas:
(a) y = x2 − 2x − 2
(b) y = −x2 + 4x − 5
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Ms H.L. Tarr
Examples of using sketches of parabolas to solve inequalities
(1) For what values of x will x2 − 3 > 0?
Solution:
(2) Find all real values of x which satisfy x2 + x − 1 ≤ 0
Solution:
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Ms H.L. Tarr
Tutorial 2: Intervals of the Real Line & Quadratics
(1) Express in interval form (N.B. ”,” means ”and”)
(a) x < 0 and x < 2
(b) x < 0, x ≥ 2
(c) x ≥ −2, x ≤ 1
(d) x < 2 or x ≥ 0
(e) x ∈ (0, 2) or x ≥ 2
(f) x ≥ 2 or x < 1
(g) 0 ≤ x ≤ 2 or x ∈ ∅ (h) x ∈ [0, 2] and x ∈ (1, 3) (i) x ≥ −2, x < 3
(j) x ≤ −2, x > 3
(k) x ≤ −3 and x ∈ R
(m) x < 3 or x < 7
(l) x ≤ −3, x > −
11
3
(2) Solve the following equations:
(a) 9x(x − 1) = −2 (b) 3x2 = 4x(x + 1) (c) 7x2 = 3(x3 + x) (d) 3x2 + x = −1
(3) If p is a solution of 3x2 − 5x + 5 = 0, find the value of 3p2 − 5p − 9.
(4) Given x2 + 5x + 6 = 0, use the discriminant to determine whether the roots of this
equation are real/not real, rational/irrational, equal/unequal. Give reasons.
(5) If x2 −6x+k = 0, for what value of k will the roots be: (a) equal (b) real (c) imaginary ?
(6) Show that x2 + (a + b)x + ab has rational roots for a, b ∈ Q.
2
(7) Show that x2 − bx + b2 > 0 ∀x ∈ R and non-zero b ∈ R. (Hint: show that the graph
2
of x2 − bx + b2 does not cut the x-axis i.e. lies above the x-axis).
Tutorial 2 Solutions
x ∈ (−∞, 0) (b) x ∈ ∅
x ∈ (0, ∞)
(f) x ∈ (−∞, 1) ∪ [2, ∞)
x ∈ [−2, 3)
(j) x ∈ ∅
x ∈ (−∞, 7)
√
7 ± 13
1 2
,
(b) 0, −4 (c) 0,
(d)
(2) (a)
3 3
6
(1)
(a)
(e)
(i)
(m)
(c) x ∈ [−2, 1]
(d) x ∈ R
(g) x ∈ [0, 2]
(h) x ∈ (1, 2]
(k) x ∈ (−∞, −3] (l) x ∈ (− 11
, −3]
3
no real solutions
(3) 3p2 − 5p − 9 = −14
(4) Roots are real (∆ > 0), rational (∆ is a perfect square) and unequal (∆ 6= 0).
(5) (a) k = 9 (b) k ≤ 9 (c) k > 9
(6) Use quadratic formula to find roots, or show that ∆ is a perfect square.
(7) ∆ < 0 since b2 > 0, and a = 1 > 0 i.e
not cut the x-axis.
S
12
, so the graph lies above the x-axis and does
Ms H.L. Tarr
1.5 Inequalities
N.B. This is important for finding domains of functions.
We use the following basic rules, where a, b, c ∈ R
(1) For a, b ∈ R, either a < b or a = b or a > b.
(2) a < b and b < c ⇒ a < c (Transitive Law).
(3) a < b ⇒ a + c < b + c for any a ∈ R
(4) a < b and c > 0 ⇒ ac < bc
a < b and c < 0 ⇒ ac > bc
(5) If a, b have the same sign, then a < b ⇒ 1/b < 1/a
Example 1: For x ∈ (−1, 1), find the interval in which 3 − 2x lies.
Solution:
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Ms H.L. Tarr
1.5.1 Solving Inequalities
(a) Basic Algebraic Method
p(x)
p(x)
< 0 or
> 0 by taking everything over to the LHS, then
q(x)
q(x)
simplifying and factorising.
(ii) Find all the zeros of the numerator and denominator (the function is undefined at zeros
of the denominator).
(iii) Arrange all these zeros in increasing order on a line.
(iv) Work out the sign on each interval using the quick method described below (or a table of
signs).
(v) Check the endpoints of each interval in your answer.
(i) Express in the form
Method - signs on intervals (or a table of signs)
• Note that x − α (or α − x) changes sign at x = α (where it becomes zero). So does any
odd power of x − α, but not an even power of x − α.
• Draw a coordinate line, using the values where x = α to divide the line into intervals.
• Work out the sign for the function in the interval to the right of the greatest zero (i.e.
highest value of α) by using a test value for x in that interval.
• Proceed to the left, changing or not changing the sign at each of the zeros, depending
on the power of the factor.
N.B NEVER divide an inequality by a variable!
Example 1: Solve 2 − x2 ≥ 0
Solution:
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Ms H.L. Tarr
Example 2: Solve
Solution:
x(x + 2)2 (1 − 3x)
≤0
(x − 2)(x2 + 4)
Exercise 8 Solve x2 > 4x − 4 for x
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Ms H.L. Tarr
(b) Graphical Method
Example 1: Solve for x: x ≥
1
x
Solution:
Exercise 9 Show graphically that the solution set for the inequality x2 − 3x > 10 is
(−∞, −2) ∪ (5, +∞)
Solution:
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Ms H.L. Tarr
Tutorial 3: Inequalitites
(1) Use a rough sketch of a graph to determine the values of x ∈ R for which the following
inequalities hold:
(a) (x − 1)(x + 4) < 0 (b) (2 − x)(x + 3) < 0 (c) x2 − 4 > 0
(d) x2 ≤ 9
(e) 3 − x2 ≥ 0
(f) x2 − 2x − 2 ≥ 0
(h) x2 − 4x + 2 ≤ 0
(i) x2 > 8
(g) x2 − 2x + 2 > 0
(2) True or False? (Justify your answer!)
x2 − 2xy + y 2 ≥ 0 for all x, y ∈ R
(3) Prove the following theorem graphically:
For a > 0 (a) x2 < a2 ⇒ x ∈ (−a, a)
(b) x2 > a2 ⇒ x ∈ (−∞, −a) ∪ (a, ∞)
(4) Solve for x, expressing your answer in interval form.
(x − 1)(x + 2)
x(x − 1)
(a) −2 < 1 − 3x < 5 (b)
>0
≤ 0 (c)
2
(x + 1)
(x2 + 1)
x
x(x − 4)
(e)
≤0
(f)
(d) x(x − 3)4 ≤ 0
≥0
2
1−x
2+x
1
(5) Show that, for 0 ≤ x ≤
2
2
2
1
16
1
1
≤ 1 (c)
≤1
(a)
≤
≤ 1 (b)
≤
<√
4
3
x+1
17
1+x
5
x+1
√
3
(use 2 < )
2
(6) If x ∈ (1, 9), determine the intervals in which the following functions lie:
√
1
1
(a) √ + 2 (b)
x + 3 − 3 (c)
x
(x − 1)1 /3 + 4
Tutorial 3 Solutions
(1) (a)
(c)
(e)
(g)
(i)
x ∈ (−4, 1)
(b) x ∈ (−∞, −3) ∪ (2, ∞)
x ∈ (−∞,
−2)
∪
(2,
∞)
(d)
x ∈ [−3, 3]
√ √
√
√
x ∈ [− 3, 3]
(f) x ∈ (−∞,√1 − 3]√∪ [1 + 3, ∞)
x∈R
(h) x ∈ [2 − 2, 2 + 2]
√
√
x ∈ (−∞, −2 2) ∪ (2 2, ∞)
(2) True. Why?
(4) (a) x ∈ (− 43 , 1)
(b)
(c) x ∈ (−∞, 0) ∪ (1, ∞) (d)
(e) x ∈ (−1, 0] ∪ (1, ∞)
(f)
√
7
(6) (a) ( 3 , 3) (b) (−1, 2 3 − 3)
x ∈ [−2, −1) ∪ (−1, 1]
x ∈ (−∞, 0] ∪ {3}
x ∈ (−2, 0] ∪ [4, ∞)
(c) ( 16 , 14 )
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Ms H.L. Tarr
Tutorial 4: Solving Inequalities Graphically
(1) From the following graphs, read off the values of x for which f (x) < g(x), expressing
answers in interval form.
(a)
(b)
(c)
(2) Solve the following inequalities graphically,
i.e. in each case:
(a) Sketch the function on the LHS of the inequality and the function on the RHS of
the inequality on the same set of axes,
(b) Find the values of x for which the two graphs intersect,
(c) Read off the answer from the sketch and express in interval form.
(ii) x4 ≥ x1
(i) x2 > 2
(iii) x3 < 4x (iv) x2 − 4 ≤ x − 2
Tutorial 4 Solutions
(1) (a) x ∈ (−∞, b) ∪ (c, ∞) (b) x ∈ (0, c) (c) x ∈ (a, 0) ∪ (d, ∞)
(2)
√
√
(i) x ∈ (−∞, − 2) ∪ ( 2, ∞) (ii) x ∈ [−2, 0) ∪ [2, ∞)
(iii) x ∈ (−∞, 2) ∪ (0, 2)
(iv) x ∈ [−1, 2]
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Ms H.L. Tarr
1.6 Absolute Value or Modulus(”makes things positive”)
N.B
Definition 1.10. For a ∈ R, |a| =
a,
for a ≥ 0
−a, for a < 0
e.g. |2| = 2
| − 5| = −(−5) = 5 since −5 < 0
(Note that for a < 0, −a > 0 so that |a| ≥ 0 ∀ a ∈ R)
N.B. Similarly, for functions,
f (x),
for f (x) ≥ 0
|f (x)| =
−f (x), for f (x) < 0
e.g.
(i) For x − 2 ≥ 0, i.e. x ≥ 2, |x − 2| = x − 2
(ii) For x + 1 < 0, i.e. x < −1, |x + 1| = −(x + 1)
(iii) If x > 1, x − 1 > 0 so that |x − 1| = x − 1
(iv) If x < −2, x + 2 < 0 so that |x + 2| = −(x + 2)
(v) If x < 1/2, 2x − 1 < 0 so that |2x − 1| = −(2x − 1) = −2x + 1
N.B.Properties of Absolute Value
(1) |f (x)| ≥ 0 for any function f (x)
(2) |f (x)| = | − f (x)|
e.g. |x − 2| = |2 − x|
(3) |x + y| ≤ |x| + |y|
(Triangle Inequality)
(4) |xy|
√ = |x||y|
√
(5)
x2 = |x|
( x2 6= x for x < 0)
e.g. |4 − x2 | = |(2 − x)(2 + x)| = |2 − x| |2 + x| = [−(2 − x)](2 + x) for x > 2
N.B.
Theorem 1.11.
For b > 0, |x| = b ⇔ x = ±b
|x| < b ⇔ −b < x < b
|x| > b ⇔ x < −b or x > b
Sketching graphs involving modulus
Recap: Straight line graph y = mx + c
where m = gradient/slope (”rise over run”)
and c = y−intercept
Quick method for plotting: Join the x− and y−intercepts (if c 6= 0). If c = 0, you will need to
plot one other point, since the graph passes through the origin.
Check :
If m < 0 ⇒
If m > 0 ⇒
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Ms H.L. Tarr
Recall:
• Two straight lines are parallel if m1 = m2
• Two straight lines are perpendicular if m1 · m2 = −1
N.B.The graph of y = |f (x)| for any function f (x)
Method:
(i) Draw the graph of y = f (x) in pencil.
(ii) Do not change the portion of the graph that lies above the x−axis (this is where the
function has positive values i.e. y ≥ 0).
(iii) Reflect the part of the graph that lies below the x−axis about the x−axis i.e. draw the
mirror-image of the part of the graph that lies below the x−axis (this portion of the graph
is where the function has negative values i.e. y < 0).
This will be the graph of y = −f (x).
Example 1: Sketch the graph of y = |2x − 1|
Solution:
Example 2: Sketch the graph of y = |(x − 1)(x + 2)|
Solution:
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Ms H.L. Tarr
Example 3: Sketch the graph of y =
1
1
=
|x|
x
Solution:
Solving equations or inequalities involving modulus
Method 1: Use the Theorem 1.11 for problems of the type
|f (x)| < α or |f (x)| = α or |f (x)| > α where α is positive.
Example 4: |1/x − 1| < 1
Solution:
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Ms H.L. Tarr
Method 2: Graphical - this method is often simpler and ”safer” than using an
algebraic method.
Note: draw the graph roughly to scale and make sure that the lines don’t
look parallel when they are not!
e.g. Solve |3x − 1| ≤ |2 − x| i.e. |3x − 1| ≤ |x − 2| ............ (a)
Solution:
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Ms H.L. Tarr
Method 3: Algebraic. We use the definition of absolute value (pg 19) to remove the
modulus sign, considering 2 cases separately.
e.g. Solve the inequality x ≤ |x − 1|.
Case 1
Case 2
Exercise 10 What if x > |x − 1| ?
The algebraic method becomes complicated
- for 2 or more factors under the modulus sign e.g. |x2 − x| < x + 1; or
- for 2 separate modulus signs e.g. |3x − 1| < |x + 2|
- in cases like these, use the graphical method instead.
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Ms H.L. Tarr
Method 4: Distance Interpretation. Uses the fact that |x − a| is the distance of x from a.
(Visualize a number-line: more on this concept later).
Sketching graphs involving more than one modulus
Example 1: y = |x + 1| − |x − 4|
Exercise 11 Plot the graph from the above example.
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Ms H.L. Tarr
Tutorial 5: Absolute Value
(1) Use the definition of modulus to complete the following, simplifying where possible.
N.B. Justify your answers.
(a) For x > 5, |x − 5| = ?
(b) For x > 2, |2 − x| = ?
(c) For x < 2, |2x − 4| = ?
(d) For x < − 13 , |3x + 1| = ?
2|
|x2 −2x|
=?
(e) For 0 < x < 2, x−2 = ? *(f) For x > 12 , |1−4x
x− 1
2
(2) Solve the following inequalities, using the theorem (or a distance argument).
(a) |x − 3| < 5
(b) |4x − 1| ≤ 1
(d) |3x − 4| > 2
(c) |x − 12 | ≥ 1
(e) 0 < |x − 2| < 1
(f) 0 < |x − c| < δ
(g) |y − M | < ε
(h) |x − 2| ≥ −3
(i) |x2 + 2| < |x2 + 1| *(j) 1 < |x + 1| ≤ 2
*(k) |x − 1| < |2 − x|
(l) |x − 3| ≤ |x + 1|
(3) For what values of x, in interval form, is f (x) ≤ g(x)?
(a)
(b)
(4) Solve the following inequalities graphically:
(a) |x − 3| ≤ 5
(b) |3x − 4| > 2
(c) |x − 3| ≤ |x + 1| (d) |x − 1| < |2 − x|
*(e) |x2 − 3| < 1
(5) Sketch the graph of y = |f (x)| if the graph of y = f (x) is as shown.
(a)
(b)
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Ms H.L. Tarr
(c)
(d)
(6) Draw quick sketches of the following functions:
(a) y = |3x − 6| (b) y = |x + 2| (c) y = |x2 |
1
(d) y = |1 − x2 | (e) y = |x3 |
(f) y =
x
Tutorial 5 Solutions
(1) (a) x − 5
(2) (a)
(c)
(e)
(g)
(i)
(k)
(b) x − 2
(c) 4 − 2x
(d) −3x − 1
(e) −x
(f) 4x + 2
x ∈ (−2, 8)
(b) x ∈ [0, 12 ]
1
3
x ∈ (−∞, − 2 ] ∪ [ 2 , ∞) (d) x ∈ (−∞, 23 ) ∪ (2, ∞)
x ∈ (1, 2) ∪ (2, 3)
(f) x ∈ (c − δ, c) ∪ (c, c + δ)
y ∈ (M − ε, M + ε)
(h) x ∈ R
x∈Q
(j) x ∈ [−3, −2) ∪ (0, 1]
3
(l) x ∈ [1, ∞)
x ∈ (−∞, /2)
(3) (a) x ∈ [0, ∞) (b) x ∈ ∅
(4) (a) x ∈ [−2, 8]
(b) x ∈ (−∞, 23 ) ∪ (2, ∞)
(c) x ∈ [1, ∞) √
(see 2l) √
(d) x ∈ (−∞, 32 ) (see 2k)
(e) x ∈ (−2, − 2) ∪ ( 2, 2)
(5)
(a)
(b)
(c)
(d)
(e)
(f)
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Ms H.L. Tarr
Tutorial 6: Absolute Value Worksheet
Definition:
(
a if a ≥ 0
−a if a < 0
if a is any real number
|a| =
(
x+5
if x + 5 ≥ 0
Examples: |x + 5| =
−(x + 5) if x + 5 < 0
(
x+5
if x ≥ −5
=
−x − 5 if x < −5
−−−−−−−−−−−−−−−−−→
Method 1 - Use the definition
|x − 5| = 3
∴ x − 5 = 3 if (x − 5) ≥ 0 or −(x − 5) = 3 if (x − 5) < 0
∴ x = 8 if x ≥ 5
or ∴ x = 2 if x < 5
−−−−−−−−−→
−−−−−−−−−→
Method 2 - Square both sides
|x − 5| = 3
2
∴
|x − 5| = (3)2
∴ x2 − 10x + 25 = 9
∴ x2 − 10x + 16 = 0
∴ (x − 8)(x − 2) = 0
∴ −
x−=
or−−
x−=
−−8−−
−→2
N.B. Check roots !
x = 8 : LHS = |8 − 5| = |3| = 3 = RHS
x = 2 : LHS = |2 − 5| = | − 3| = 3 = RHS
Exercises
(1) Solve for x:
(a) |2x − 7| = x + 3 (b) |x − 3| = 7
(c) |x + 9| = −5
(d) |2 − x| + 3 = 0
1
(e) |x − 1| = 2 x + 1 (f) |x − 1| = 2x + 1
x−5
(g) |x + a| = b
(h) |x − 1| =
3
(i) 2|3x − a| = 4b
(2) Draw a sketch graph of the following functions
(a) y = |x − 2|
(b) y = −|x + 2|
(c) y = 3|x − 1|
(3) (a) Solve the equation |x − 2| = 3x − 2
(b) Verify the solution by plotting y = |x − 2| and y = 3x − 2 on the same system of
axes
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Ms H.L. Tarr
Tutorial 6 Solutions
Absolute Value Worksheet
x = 10 or x = 4/3
(b) x = 10 or x = −4
No solution (x ∈ ∅) (d) No solution (x ∈ ∅)
x = 4 or x = 0
(f) x = 0
x = −a ± b
(h) No solution (x ∈ ∅)
a ± 2b
(i) x =
3
(1) (a)
(c)
(e)
(g)
(2)
(a)
(b)
(c)
(3) (a) x = 1
(b)
NOTE:
Only one point of
intersection P
i.e. only one root of
|x − 2| = 3x − 2
Tutorial 7: Absolute Value Worksheet 2
For absolute value equations:
If |x| = a, then x = ±a (provided a ≥ 0) ... (1)
* Remember to check roots
For absolute value inequalities:
If |x| < a, then −a < x < a
If |x| ≤ a, then −a ≤ x ≤ a (provided a ≥ 0) ... (2)
If |x| > a, then x < −a or x > a
If |x| ≥ a, then x ≤ −a or x ≥ a (a ∈ R) ... (3)
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Ms H.L. Tarr
Examples
(1) |3x + 1| − 5 = 0 ∴
|2x + 1| = 5
∴
2x + 1 = ±5 (using (1))
∴
2x + 1 = 5
or 2x + 1 = −5
∴
2x = 4
or 2x = −6
∴
x
or x = −3
−−=
−→2
−−−−→
Check roots: x = 2 : LHS= |5| − 5 = 0 =RHS
x = −3 : LHS= | − 6 + 1| − 5 = 5 − 5 = 0 =RHS
(2) |x − 3| ≤ 7 ∴
∴
−7 ≤ x − 3 ≤ 7 (Using (2))
−4 ≤ x ≤ 10
−−−−−−−−−→
(3) |x + 1| > 3 ∴
∴
x + 1 > 3 or x + 1 < −3 (using (3))
x>2
or x < −4
−−−→
−−−−→
Exercises
(1) Solve for x and verify your solution graphically
(a) |2x − 1| < 3
(b) |x + 2| ≥ 4
(2) Solve for x:
12
−1
(a) 7 − |x − 5| > 0 (b) |x−3|
≥ 6 (c) |x−2|
≥ 0 (d)
(3) Write the following in the form |x − p| ≤ q:
(a) 1 ≤ x ≤ 7 (b) −4 ≤ x ≤ 0
(4) Write the following in the form |x − p| > q:
(a) x > 4 or x < 2
29
−2
≤5
|4−2x|
Ms H.L. Tarr
Tutorial 7 Solutions
Absolute Value Worksheet
(1) (a)
∴ −3 < 2x − 1 < 3
∴ −2 < 2x < 4
∴ −1 < x < 2
−−−−−−−−→
(b)
∴ x + 2 ≥ 4 or x + 2 ≤ −4
∴ x≥2
or x ≤ −6
−−−→
−−−−→
(2) (a) ∴
∴
∴
∴
−|x − 5| > −7
|x − 5| < 7
−7 < x − 5 < 7
−2 < x < 12
−−−−−−−−→
(b) ∴
∴
∴
∴
−1 |x − 2|
No
solution
−−−−
−−−−→
(d)
12 ≥ 6|x − 3|
|x − 3| ≤ 2
−2 ≤ x − 3 ≤ 2
−1 ≤ x ≤ 5 and x 6= 3
−−−−−−−−→
−−−→
−2 ≤ 5|4 − 2x|
∴ x ∈ R , x 6= 2
−−−−−−−−−→
(3) (a)
|x − p| ≤ q
(b)
|p − q| = −4 and p + q = 0
∴ −q ≤ x − p ≤ q
∵ p − q = −4
∴ p−q ≤x≤p+q
∵ p+q =0
∴ p − q = 1 and p + q = 7
2p = −4
∴ p = −2
p−q =1
q=2
p+q =7
∴ |x + 2| ≤ 2
−−−−−−−→
2p = 8 ⇒ p = 4 & q = 3
∴ |x − 4| ≤ 3
−−−−−−−→
(4) (a) |x − p| > q ⇒ x − p > q or x − p < −q ⇒ x > p + q or x < p − q
∴ p+q =4
∴ p=3
q = 1 ∴ |x − 3| > 1
p−q =2
−−−−−−−→
2p = 6
(c)
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Ms H.L. Tarr
1.7 Functions
Definition 1.12. A function is a correspondence, governed by a rule, between two sets, called
the domain and the range, such that for each value in the domain, there corresponds exactly
one value in the range.
i.e. a function is any rule that assigns to each element in one set some element from another
set.
Note:
(i) Each element in the domain can map onto only one element of the range.
e.g.
(ii) It is possible for a function to have two domain elements mapping onto the same range
element.
e.g.
(iii) It is not possible for a function to have one domain element mapping onto more than one
range element.
e.g.
Notation
(i) By convention, we denote a function using a small letter, such as f, g, h, etc.
e.g. f (x) = 2x is read ”f of x equals 2x” i.e the function f maps the domain value x onto
the range value 2x.
(ii) For example, the function defined by the equation, y = x2 + 2x may be referred to as
f , and written in the form f (x) = x2 + 2x, where x is the domain value and f (x) is the
corresponding range value.
Definition 1.13. The domain of a function f (x), written Dom f or Df , is defined by:
Dom f = {x ∈ R | f (x) is defined}
e.g. given the function f (x) =
1
x2 − 1
,
Dom f = {x ∈ R | x 6= ±1} = (−∞, −1) ∪ (−1, 1) ∪ (1, ∞)
N.B
p
g(x) to be defined, we must have g(x) ≥ 0.
1
to be defined, we must have g(x) > 0.
(ii) For p
g(x)
(i) For
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Ms H.L. Tarr
Example 1: Find the values of x for which the function f (x) =
√
4 − x2 is defined.
Solution:
Vertical Line Test for Functions
To each x in its domain, a function assigns exactly one y−value. Thus a straight line parallel
to y−axis (i.e. a vertical line) cuts the graph of a function at most once.
e.g
Definition 1.14. The range of a function f (x), written Rf , is defined by:
Rf = {y | y = f (x) for some x ∈ Df }
√
e.g. f (x) = 4 − x2 then Rf =..........................
It helps to sketch the function when finding the range, or to think what the largest and smallest
values of the function are.
Exercise 12 What are the largest and smallest values of the following functions?
(a) y =
1
x2
(b) y =
√
(c) y =
1
x2 + 1
x
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Ms H.L. Tarr
1.8 Composite Functions(”function of a function”)
Definition 1.15. For functions f (x) and g(x), the composition of f with g, denoted f ◦ g
(read ”f of g”) is defined by:
(f ◦ g)(x) = f g(x)
where Dom f ◦ g = {x ∈ R | g(x) is defined, f g(x) is defined}
= {x ∈ R | x ∈ Dg , g(x) ∈ Df }
Note that, in general, f ◦ g 6= g ◦ f
Composition of a function means putting two or more functions together in a special way to
produce another function.
e.g. if we have two functions,
f (x) = 2x and g(x) = x + 3, we can work out two composite
functions (f ◦ g)(x) = f g(x) and (g ◦ f )(x) = g f (x) . Consider the function diagrams below:
For g f (x) : First we apply the function f to x, and then we apply the function g to the
answer.
For f g(x) : First we apply the function g to x, and then we apply the function f to the
answer.
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Ms H.L. Tarr
Example 1: Evaluate (f ◦ g)(x) and (g ◦ f )(x), and the domains of these functions if:
√
1
f (x) =
and g(x) = x
x−4
Solution:
Example 2: Find the rule and domain of (g ◦ f )(x) if:
√
1
f (x) = and g(x) = x2 − 4
x
Solution:
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Ms H.L. Tarr
Tutorial 8: Functions: domain, range and composition
(1) Find the domain and range of the following functions. (Give your answers in interval
form).
√
1
(a) f (x) =
(b) f (x) = |x + 3| − 1 (c) f (x) = 7 − x2
x−3
√
1
1
(d) f (x) = x2 + 2x + 10 (e) f (x) = √
(f) f (x) = √ + 1
x+1
x
(2) Find the rule of the functions f ◦ g and g ◦ f , and evaluate the domain of each in interval
form.
√
1
1
(b) f (x) = and g(x) = 9 − x2
(a) f (x) = x + 2 and g(x) =
x
x
√
√
1
1
(d) f (x) = x − 2 and g(x) =
(c) f (x) = x and g(x) =
x+2
x−2
(3) Express the following functions as compositions of 2 or more functions
√
1
(a) f (x) = (3 − x2 )2
(b) f (x) =
(c) f (x) = 4 − 3x2
5 − 2x
p
√
(d) f (x) = |x3 − 1|
(e) f (x) = (1 − 1 + x2 )2 (f) f (x) = 1 + 3(x − 1)2
(g) f (x) = 4 + |2 + x2 |
√
(4) If f (x) = x and g(x) = x + 1, find the rule and domain of:
(a) (f ◦ g)(x) (b) (g ◦ f )(x) (c) (f ◦ f )(x) (d) (g ◦ g)(x)
Tutorial 8 Solutions
(1)
(a)
(b)
(c)
(d)
(e)
(f)
(2)
(a)
(b)
(c)
(d)
(3)
(4)
Df = (−∞, 3) ∪ (3, ∞),
Df = R,√ √
Df = [− 7, 7],
Df = R,
Df = [0, ∞),
Df = (0, ∞),
1
(f ◦ g)(x) = + 2,
x
1
(g ◦ f )(x) =
,
x+2
1
,
(f ◦ g)(x) = √
2
r9 − x
9x2 − 1
,
(g ◦ f )(x) =
x2
1
(f ◦ g)(x) = √
,
x
r +1
5 − 2x
,
(f ◦ g)(x) =
x−2
Rf = (−∞, 0) ∪ (0, ∞)
Rf = [−1,
√∞)
Rf = [0, 7]
Rf = [3, ∞) (Hint: Complete the square)
Rf = (0, 1]
Rf = (1, ∞)
Df ◦g = (−∞, 0) ∪ (0, ∞)
Dg◦f = (−∞, −2) ∪ (−2, ∞)
Df ◦g = (−3, 3)
Dg◦f = (−∞, − 31 ] ∪ [ 31 , ∞)
Df ◦g = (−2, ∞)
Df ◦g = (2, 25 ]
1
, Dg◦f = [0, ∞)
x+2
1
(g ◦ f )(x) = √
, Dg◦f = [2, 6) ∪ (6, ∞)
x−2−2
(g ◦ f )(x) = √
f (x) = (g ◦ h)(x) where g(x) = x2 and h(x) = 3 − x2
1
and h(x) = 5 − 2x
f (x) = (g ◦ h)(x) where g(x) = √
x
f (x) = (g ◦ h)(x) where g(x) = x and h(x) = 4 − 3x2
f (x) = (g ◦ h)(x) where g(x) = |x| and h(x) = x3 −
√1
f (x) = (g ◦ h ◦ j)(x) where g(x) = x√2 , h(x) = 1 − x and j(x) = 1 + x2
f (x) = (g ◦ h ◦ k)(x) where g(x) = x, h(x) = 1 + 3x2 and k(x) = x − 1
f (x) = (g ◦ h)(x) where g(x) = 4 + |x| and h(x) = 2 + x2
√
√
1
x + 1, Df ◦g = [−1, ∞) (b)
x + 1, Dg◦f = [0, ∞) (c) x 4 , Df ◦f = [0, ∞)
(a)
(d) x + 2, Dg◦g = R
(a)
(b)
(c)
(d)
(e)
(f)
(g)
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Ms H.L. Tarr
1.9 Symmetry (Even and Odd Functions)
Definition 1.16. A function f (x) is even if: f (−x) = f (x) for all x ∈ Df .
i.e. the graph is symmetric about the y−axis
• Since f (−a) = f (a), the graph has the same y−value at x = a and x = −a.
Examples of even functions are 4, x2 , 1 + x2 , cos x
Example 1: Is f (x) = x2 + 3x + 2 an even function ?
Solution:
Example 2: Is f |x| an even function for any function f (x) ?
Solution:
Definition 1.17. A function f (x) is odd if: f (−x) = −f (x) for all x ∈ Df .
i.e. the graph is symmetric about the origin 0
(Since f (−a) = −f (a), for any point P = a, f (a) on the graph, the point Q = − a, f (−a) =
− a, −f (a) is also on the graph).
Examples of odd functions are x, 1/x for x 6= 0, x3 , sin x.
Example 3: Show that f (x) is an odd function if f (x) =
x3
.
1 + x2
Solution:
Note: When sketching an odd or even function, we need only consider x ≥ 0 and we can then
draw in the rest of the graph by symmetry.
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Ms H.L. Tarr
Example 4: Given y = f (x) for x ≥ 0:
For f even:
For f odd:
Properties:
These help us to decided quickly whether a function is even or odd, but using them is not a proof!
A proof must use the definitions above.
even • even is even
even ± even is even
odd • odd is even
odd ± odd is odd
even • odd is odd BUT even ± odd is NEITHER odd nor even
Exercise 13 Show that if f (x) is an odd function and g(x) is an even function, then f g is
odd.
Solution:
37
Ms H.L. Tarr
Tutorial 9: Symmetry
(1) If the graph of y = f (x) is as shown, draw the graphs of:
(i) y = −f (x) (ii) y = f (−x) (iii) y = f (x) − 1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Tutorial 9 Solutions
(1) (a)
(b)
(i)
(ii)
(iii)
(i)
(ii)
(iii)
38
Ms H.L. Tarr
(c)
(d)
(e)
(f)
(g)
(i)
(ii)
(iii)
(i)
(ii)
(iii)
(i)
(ii)
(iii)
(i)
(ii)
(iii)
(i)
(ii)
(iii)
Note: H.A. = horizontal asymptote, V.A. = vertical asymptote
39
Ms H.L. Tarr
Tutorial 10: Symmetry 2
(1) Determine whether the following functions are even, odd or neither.
x
(b) g(x) = (1 − x)2 (c) h(x) = x|x|
(a) f (x) = 2
x −1
1 − x2
(e) h(x) = x5 − 3x3 (f) f (x) = x5 − 3x3 + 1
1 + x2
(2) Prove
that
if
f
(x)
is
odd
and
g(x)
is
even,
then
the
function
h(x)
=
f
(x)
−
g(x)
f (x) +
g(x) is even.
(d) g(x) =
(3) Prove that f (0) = 0 if f (x) is an odd function defined at x = 0.
(4) Can the graph of a function be symmetric about the x-axis?
(5) Prove that if g(x) is even, f ◦ g is even for any function f (x).
(6) Complete the following graphs, if possible, to make the function: (a) odd (b) even
(a)
(b)
(c)
(d)
(e)
(f)
Tutorial 10 Solutions
(1) (a) odd (b) neither (c) odd (d) even (e) odd (f) neither
(4) No. A vertical line will cut a curve that is symmetric about the x-axis in at least 2
points, so the curve cannot be the graph of a function.
40
Ms H.L. Tarr
(6)
(a)
(b)
(c)
(d)
(e)
(f)
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Ms H.L. Tarr
1.10 Translations of Functions (Graphs from known graphs)
1.10.1 Graphs of basic functions
You should know the graphs of the following basic functions:
Some families of graphs have the same basic shape, and a specific graph of a function can
be obtained by ”decomposing” the given function into its composite functions and using the
standard graphs. The changes are known as ”translations” of the original graph.
1.10.2 Basic Types of Translations (c > 0)
(1) Shifts
Original graph
• Horizontal shift c units to the right
• Horizontal shift c units to the left
• Vertical shift c units down
• Vertical shift c units up
:
:
:
:
:
y = f (x)
y = f (x − c)
y = f (x + c)
y = f (x) − c
y = f (x) + c
(2) Reflections
Original graph
• Reflection about the x−axis
• Reflection about the y−axis
• Reflection about the origin
:
:
:
:
y = f (x)
y = −f (x)
y = f (−x)
y = −f (−x)
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Ms H.L. Tarr
Example 1: Original graph y = x2
Example 1 illustrates the 3 basic types of translations, viz. vertical shifts, horizontal shifts
and reflections. Function notation is useful for describing translations of graphs in the plane.
For example, for y = x2 as above, the translations shown can be represented by the following
equations:
y = f (x) + 2
y = f (x + 2)
y = −f (x)
y = −f (x + 3) + 1
Vertical shift up 2 units
Horizontal shift 2 units to the left
Reflection about the x−axis
Shift left 3 units, reflect about x−axis, shift up 1 unit.
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Ms H.L. Tarr
1.10.3 Summary of Translations
(1) y = f (x) ± c for a positive constant c (Vertical Translation)
i.e. for each x, the y−value increases or decreases by the same amount c.
So, for ”+” sign, move graph c units up.
for ”−” sign, move graph c units down.
(2) y = f (x ± c) for a positive constant c (Horizontal Translation)
i.e. the graph shifts either left (”+” sign) or right (”−” sign) a distance of c units. e.g.
(3) y = −f (x) (Reflection about x−axis)
i.e. For each x−value, the y−value changes sign. e.g.
Form the mirror image about the x−axis.
(4) y = f (−x) (Reflection about y−axis)
i.e. The y−value at x becomes the y−value at
−x and vice versa. Let g(x) = f (−x). Then,
g(0) = f (0).
g(a) = f (−a)
g(−a) = f −(−a) = f (a)
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Ms H.L. Tarr
Example 1:
(a) Sketch y = (x − 2)3
(b) Sketch y = | − x|
(c) Sketch y =
√
x+2
Exercise 14 Sketch the above graphs.
Combinations of operations
N.B. Sketch and label your original graph, and each step.
Example 2: Sketch the graph of y = 1 − x1 from the graph of y = x1 .
g(x) = −f (x)
f (x)
(a)
− x1 + 1
:
or
(b)
1
x
→
f (x)
1
1 + −x
:
or
(c) −( x1 − 1) :
1
x
h(x) = g(x) + 1
→
g(x) = f (x) + 1
1
+1
x
→
1
−1
x
→
− x1 + 1
h(x) = g(−x)
→
g(x) = f (x) − 1
f (x)
1
x
− x1
1
1 + −x
h(x) = −g(x)
→
−( x1 − 1)
Exercise 15 Sketch y = 1 − x1 using each of the above approaches (a), (b) and (c).
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Ms H.L. Tarr
Tutorial 11: Translation of Functions
(1) In each of the following, express g(x) in terms of f and sketch its graph from
the graph of f (x).
, g(x) = (x + 2)2
(a) f (x) = x2
(b) f (x) = x2
, g(x) = −x2
(c) f (x) = |x|
, g(x) = |x| − 1
(d) f (x) = |x|
, g(x) = |x − 1|
(e) f (x) =
, g(x) =
1
x
1
(f) f (x) =
x
√
(g) f (x) = x
*(h) f (x) = ex
*(i) f (x) = lnx
1
+1
x
1
, g(x) =
x+1
√
, g(x) = −x
, g(x) = ex − 1
, g(x) = ln(−x)
(j) f (x) = x2 + x , g(x) = |x2 + x|
(2) Draw quick sketches of the following graphs from standard graphs. Put in intermediate steps
and label any intercepts and asymptotes which occur.
√
(a) y = − x + 1 (b) y = |(x − 1)3 | (c) y = 1 − x3 *(d) y = ln|x| *(e) y = −e−x
Note: Do these examples after the section covering Natural Logarithmic and Exponential Functions.
Tutorial 11 Solutions
(1)
(a) g(x) = f (x + 2)
(d) g(x) = f (x − 1)
(b) g(x) = −f (x)
(e) g(x) = f (x) + 1
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(c) g(x) = f (x) − 1
(f) g(x) = f (x + 1)
Ms H.L. Tarr
(g) g(x) = f (−x)
(h) g(x) = f (x) − 1
(i) g(x) = f (−x)
(j) g(x) = |f (x)|
(2)
(a)
(d)
(b)
(c)
(e)
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Ms H.L. Tarr
1.11 Trigonometry Review
1.11.1 Degrees → Radians
Since
180◦ = π radians
1◦ = π/180 radians
⇒ x◦ = x/180 × π radians
1.11.2 Standard angles from 0 to 2π radians
Degrees 30◦ 45◦ 60◦ 90◦ 120◦ 135◦ 150◦ 180◦ 270◦ 360◦
Radians
π
6
π
4
π
3
π
2
2π
3
3π
4
5π
6
π
3π
2
2π
1.11.3 Direction of measurement
Angles are measured from the positive x-direction
with anticlockwise direction being positive.
1.11.4 Definitions
sin θ =
y
opposite
=
hypotenuse
r
cos θ =
x
adjacent
=
hypotenuse
r
tan θ =
opposite
y
=
adjacent
x
cosec θ(csc θ) =
1
r
=
y
sin θ
secantθ(sec θ) =
1
r
=
x
cos θ
cotangentθ(cot θ) =
1
x
=
y
tan θ
Recall
π π π π 2π 3π 5π
NB: Know sin θ and cos θ in surd form for θ = 0, , , , , , ,
and π
6 4 3 2 3 4 6
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1.11.5 Standard Angles; Graphs
π
6
π
4
1
2
√
3
1
2
1
√
2
π
3
√
3
2
1
√
2
1
2
θ
0
sin θ
0
cos θ
π
2
1
0
2π
3
√
3
2
−
3π
4
5π
6
π
3π
2π
2
1
√
2
1
2
√
0
−1
0
3
1
1
−√ −
−1
2
2
2
0
1
Note
(1) sin θ and cos θ repeat every 2π radians;
tan θ repeats every π radians.
(2) Since tan θ =
sin θ
, we can obtain the graph of tan θ from the graphs of sin θ and cos θ.
cos θ
(3) NB: −1 ≤ sin θ ≤ 1 ⇒ | sin θ| ≤ 1 ∀θ ∈ R
NB: −1 ≤ cos θ ≤ 1 ⇒ | cos θ| ≤ 1 ∀θ ∈ R
1.11.6 Periodicity
For n ∈ Z sin(θ + 2nπ) = sin θ
cos(θ + 2nπ) = cos θ
tan(θ + nπ) = tan θ
1.11.7 Reduction Formulae
2nd Quadrant
3rd Quadrant
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4th Quadrant
Ms H.L. Tarr
1.11.8 Signs of the Trig Ratios in the 4 Quadrants
1.11.9 Using a base angle β to solve for obtuse angle α
e.g. Solve for α where cos α = −
1
2
Method:
(1) Find a base angle β in the first quadrant by ignoring the sign initially.
i.e. cos β = 1/2 ⇒ β = π/3
(2) From the sign, determine which quadrant(s) α lies in i.e. cos α is negative ⇒ 2nd & 3rd
quads.
(3) Find α using the Reduction formulae.
or α = π + π3 = 4π
(or − 2π
)
i.e. α = π − π3 = 2π
3
3
3
1.11.10 Solving Trig Equations
NB: Learn These!!
sin x = sin α ⇒ x = α + 2nπ or x = π − α + 2nπ, n ∈ Z
cos x = cos α ⇒ x = ±α + 2nπ, n ∈ Z
tan x = tan α ⇒ x = α + nπ, n ∈ Z
(1.1)
Note: Each of these equations has at most 2 solutions in [0, 2π]. (Recall the periodicity of these
functions).
Auxiliary Angle Method
For equations of the type a cos x ± b sin x
• Convert into a single√sine/cosine function by ”forcing and compensating” i.e.
√ divide
2
2
constants a and b by a + b and compensate by multiplying entire term by a2 + b2 .
So,
√
b
a
2
2
a +b √
cos x ± √
sin x
• a cos x ± b sin x =
a2 + b 2
a2 + b 2
√
= √a2 + b2 [cos θ · cos x ± sin θ sin x]
a2 + b2 cos(θ ∓ x) by Addition Formula
=
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Ms H.L. Tarr
i.e.
a cos x ± b sin x =
√
a2 + b2 cos(θ ∓ x) =
√
a2 + b2 sin(α ± x)
(Note: To get sin instead of cos, use angle α instead)
This method enables us to:
(1) Find the max and min values of A = a cos x ± b sin x, since for n ∈ Z
Amax =
√
Amin =
√
a2 + b2 · (1) for x = ±θ + 2nπ
a2 + b2 · (−1) for x = π ± θ + 2nπ where |cos(θ ∓ x)| ≤ 1
(2) Solve equations of type a cos x ± bsinx = C
Exercise 16 Solve the following for x using the formulae from (1.1) above.
(1) sin 3x = cos
√x*
(2) sin 3x = − 3 cos 3x
(3) sin x/2 = 2
*(Hint: Convert cos x to sin(π/2 − x) or sin 3x = cos(π/2 − 3x)
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Essential Trigonometric Identities
(N.B Learn these!)
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Ms H.L. Tarr
Example 1: Reducing obtuse angles (using Periodicity)
Solution:
Example 2: Solving for an obtuse angle
Find α if sin α = − 12
Solution:
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Ms H.L. Tarr
Example 3: Solving trigonometric equations
Solve for x:
cos(4x + π/3) = −1/2
Solution:
Example 4: Solving equations using substitution
Solve sin2 x − 2 sin x − 3 = 0 for x ∈ [−π, π]
Solution:
Example 5: Auxiliary Angle Method
Solve for x:
cos x −
√
3 sin x = −1 Type: a cos x ± b sin x = c
Solution:
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Ms H.L. Tarr
Tutorial 12: Trigonometry
(1) Simplify, leaving your answers in surd form:
(a) cos(−π/6)
(b) sin(−π/3)
(c) sin(7π/6) (d) cos(2π/3) (e) sin(5π/6)
31π
4π
(g) cos( /3)
(h) sin(23π/4) (i) sin(3π/2) (j) cos(7π/4)
(f) cos( /6)
π
π
(k) tan(π + /3) (l) cot(π − /6)
(2) Find all angles√α ∈ (−π, π] such that:
(a) sin α = − 3/2 (b) cos α = −1/√2
(e) sin α = 1/2
(f) cos α = −1√
sin α = 3/2
(i) tan α = 1/√3 (j)
cos α = −1/2
(c) sin α = −1
√
(g) tan α = 3
sin α = −1√
/2
(k)
− 3
cos α = /2
√
(d) cos α = − 3/2
(h) tan α = −1
(3) Find all solutions
√ of the following equations:
(a) sin x = 3 cos x
(b) sin 2u = sin u
(c) sin(3x − π/4) = 12
π
−1 √
(d) cos(4x +√/2) = / 2 (e) 2 cos α + tan α = sec α (f) 2 cos3 θ − cos θ = 0
(g) cos x − 3 sin x = 1
(4) For which value(s) of x ∈ [0, π] will the following pairs of curves intersect ?
(a) y = tan x, y = 2 sin x (b) y = cos 3x, y = cos 2x
(5) (a) Is there a real number t such that 7 sin t = 9?
(b) Is there a real number t such that 3 csc t = 1?
Explain your answers.
(6) Express:
(a) cos4 x in terms of cos 2x and cos 4x. (b) sin2 (x/2) in terms of cos x.
(7) Find (a) (f ◦ g)(π) and (b) (g ◦ f )(π) if:
(i) f (t) = cos t and g(t) = t/4 (ii) f (t) = tan t and g(t) = t/4.
(8) If θ and φ are acute angles such that csc θ = 5/3 and cos φ = 8/17, evaluate
without using a calculator:
(a) sin(θ + φ) (b) tan(θ − φ) (c) sin 2φ (d) tan 2θ (e) sin(θ/2)
Tutorial 12 Solutions
√
√
√
3
(1) (a) 3/2 (b) − 3/2 (c) −1/2 (d) −1/2 (e) 1√
/2 (f) − √
/2
−1
−1 √
1√
(g) /2 (h) / 2 (i) −1 (j) / 2 (k)
3 (l) − 3
(c) −π/2
(d) ±(π − π/6) = ±5π/6 (e) π/6, 5π/6 (f) π
(2) (a) −2π/3, −π/3 (b) ±3π/4
(k) −5π/6
(g) −2π/3, π/3 (h) −π/4, 3π/4 (i) −5π/6, π/6 (j) 2π/3
(3) (a) x = π/3 + nπ, n ∈ Z
(b) u = 2nπ or u = π/3 + 2nπ/3, n ∈ Z
(b) or* u = nπ or u = ±π/3 + 2nπ, n ∈ Z (alternative answer)
(c) x = 5π/36 + 2nπ/3 or x = 13π/36 + 2nπ/3, n ∈ Z
(d) x = π/16 + nπ/2 or x = 3π/16 + nπ/2, n ∈ Z
(e) α = π/2 + 2nπ or α = −π/6 + 2nπ or α = 7π/6 + 2nπ, n ∈ Z
(f) θ = π/2 + nπ or θ = ±π/4 + 2nπ or θ = ±3π/4 + 2nπ, n ∈ Z
(g) x = 2nπ or x = −2π/3 + 2nπ, n ∈ Z
(4) (a) x = 0, π/3, π (b) x = 0, 2π/5, 4π/5
(6) (a) 18 (3 + 4 cos 2x + cos 4x) (b) 12 (1 − cos x)
(7) (a) (i) 1/√2 (ii) 1 (b) (i) −1/4 (ii) 0
(8) (a) 84/85 (b) −36/77 (c) 240/289 (d) 24/7 (e) 1/√10
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Ms H.L. Tarr
1.12 Natural Exponential and Logarithmic Functions
1.12.1 Revision of laws of exponents and logs
Exponents
(1) am × an = am+n
am
(2) n = am−n
a
(3) (am )n = amn
1
(4) a−m = m
a√
m
(5) a /n = n am
(6) am × bm = (ab)m
am a m
(7) m =
b
b
for a constant m, n ∈ R
Logs
(1) loga xy
=
loga x + loga y
x
(2) loga
= loga x−loga y
y
(3) loga xb = b loga x
N.B. loga (x ± y) 6= loga x ± loga y
{z
}
|
cannot be split up or simplified
Recall:
So x and y are inverse functions
1.12.2 Properties of log functions
(1) Since log functions are 1 − 1,
logb M = logb N ⇔ M = N
loga N
(2) logb N =
loga b
(3) logb bx = x ∀x ∈ R
blogb x = x for x > 0
1.12.3 Notation log10 x is written log x
1.12.4 Graph of an exponential function
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Ms H.L. Tarr
Properties of y = f (x) = ax
(1) Df : x ∈ R.
(2) Rf : y ∈ (0, +∞).
(3) (0, 1) is on the curve of f (x).
(4) The x−axis is a horizontal asymptote to the curve.
1.12.5 Graph of a logarithmic function
Recall If y = f (x), the graph of f −1 (x) can be obtained from the graph of f (x) by
reflecting it about the line y = x.
Properties of y = loga x :
(1) Df : x ∈ (0, ∞)
(2) Rf : y ∈ R
(3) x−int : (1, 0)
(4) y−axis is vertical asymptote.
1.12.6 The Natural Exponential and the Natural Logarithm
Definition 1.18. The Natural Exponential
x
1
x
y = e ∀ x ∈ R where e = lim 1 +
x→±∞
x
1
x
or
e = lim (1 + x) ≈ 2.718
x→0
(proof later)
In the family of exponential functions y = bx , the base e is an irrational number that proves to
be important in calculus.
Reason for b = e, the slope of the tangent line to the curve y = ex is equal to the y−coordinate
for any point P (x, y)
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Ms H.L. Tarr
2 < e < 3 ⇒ 2x < ex < 3x
Thus:
(1) ex > 0 and e0 = 1.
(2) ex is 1 − 1 and thus has an inverse.
(3) As x → ∞, ex → ∞
(4) As x → −∞, −x → ∞
1
1
”∞
”⇒ y = 0 is a horizontal asymptote
and ex = −x → 0
e
The usual laws of exponents hold:
e.g. (ex )y = exy
ex · ey = ex+y
etc.
Notation
ex is sometimes written exp(x)
The natural logarithm ”ln x” or ”loge x” is the inverse of the natural exponential and is defined
by:
Definition 1.19. For x > 0, y = ln x ⇔ x = ey . . . y > 0, x ∈ R
Recall: f −1 (x) ”undoes” f (x) and vice versa, so we have:
(1) ln (ex ) = x ∀ x ∈ R and eln x = x for x > 0
(2)
(3) From Definition (1.18): 1 = ln e since e = e1
0 = ln 1 since 1 = e0
The usual laws of logarithms hold:
i.e. ln xy = ln x + ln y
(product property)
x
= ln x − ln y
(quotient property)
ln
y
ln xa = a ln x for a ∈ R (power property)
1
= − ln x
(reciprocal property)
∴ ln
x
N.B. ln(x ± y) 6= ln x ± ln y
| {z }
cannot be simplified or split up.
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Ms H.L. Tarr
Also, logb x =
ln x
ln b
Behaviour of exp and log functions as x → ±∞ and x → 0
lim ex = +∞
lim ln x = +∞
x→+∞
x→+∞
x→−∞
x→0+
lim ex = 0
lim ln x = −∞
Note:
(1) For a > 0, a 6= 1: ax = ex ln a ∀ x ∈ R
x
(since ax = eln a = ex ln a )
(2) y = loga x ⇔ ay = x ∀ x > 0, y ∈ R
Later, to differentiate/integrate these functions, use (1) above to express ax in terms of e, and
use (2) to express loga x in terms of ln x as follows:
y = loga x ⇔ ay = x
⇔ ln ay = ln x
⇔ y ln a = ln x
ln x
⇔y=
ln a
ln x
So loga x =
ln a
Exercise 17
Solution:
(1) Write ln |cosec x| in terms of ln x and sin x.
(2) Show that ln |cosec x − cot x| = − ln |cosec x + cot x|
√
(3) Simplify ln 3 e.
(4) Simplify e3 ln 2
(5) Solve 4e−2t = 0.01 for t.
2
(6) Solve ln = −2 for x.
x
(7) Solve ex = −1 for x.
(8) Solve 4 · 3t = 5 · 7t for t.
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Ms H.L. Tarr
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Tutorial 13: Exponents & Logs
(1) Determine whether the following functions are odd, even or neither:
(a) f (x) = |x|ex
2
(b) f (x) = x ln(x2 + 1)
ex + 1
ex
(e)
f
(x)
=
ex − 1 √
ex + 1
(g) f (x) = ln(x + 1 + x2 )
(d) f (x) =
ex + e−x
ex − e−x 1−x
*(f) f (x) = ln
1+x
(c) f (x) =
(2) Solve the following equations:
(a) ln y = 1
(b) ln x2 = ln 2x − 1 (c) 15e2x = 5 (d) 5e2x = 20e3x
(e) e2a − 3 = 2ea
(3) Draw quick graphs of the following:
√
(a) y = 1 − e−x (b) y = 2 − x (c) y = −| ln(x − 1)| (d) y = e|x|
1
(e) y = ln |x|
(f) y = x
e
(4) What are the domains of the functions in Question 1 ?
Tutorial 13 Solutions
(1) (a) even (b) odd (c) odd (d) odd (e) neither (f) odd (g) odd
(2) (a) y = e (b) x = 2e (c) x = − 12 ln 3 (d) x = − ln 4 = −2 ln 2 (e) a = ln 3
(3)
(a)
(b)
(c)
|x|
(d) e
61
=
ex if x ≥ 0
e−x if x < 0
Ms H.L. Tarr
(e) ln |x| =
(f) e−x !!
ln x
if x ≥ 0
ln(−x) if x < 0
(4) (a), (b), (e): Df = R
(c) and (d): x 6= 0 ∴ Df = (−∞, 0) ∪ (0, ∞)
1−x
> 0 ∴ Df = (−1, 1)
(f)
1+x
√
√
√
*(g) R since 1 + x2 > x2 = |x| ≥ −x, so 1 + x2 + x > 0 for all x ∈ R.
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