MATH1051 Calculus and Linear Algebra I - Lecture Workbook Solutions

How to Use This Workbook

• This workbook is designed for lectures, tutorials, and lab sessions.
• It contains exercises, definitions, and examples to be completed during lectures.
• The completed workbook serves as a study guide for assignments and exams.
• Attending lectures is crucial.
• The calculus part of the course uses "Calculus (8th edition)" by James Stewart as a reference text.
• The workbook often refers to this text for definitions, theorems, and examples.
• The linear algebra part of the course has no set text, but "Elementary Linear Algebra (11th edition)" by Howard Anton is recommended.
• Additional course information can be found on Blackboard at http://blackboard.elearning.uq.edu.au

Numbers

Number Systems

• Mathematics relies on fundamental structures like numbers, sets, shapes, vectors, and complex numbers.
• Numbers are crucial building blocks for equations and inequalities used to solve abstract and applied problems.

Common Subsets of R (Real numbers)

• R: Set of real numbers, represented by finite or infinite decimal expansions.
• N: Set of natural numbers ${1, 2, 3, …}$
• Z: Set of integers ${0, 1, -1, 2, -2, 3, -3, …}$
• Q: Set of rational numbers, expressible in the form $\frac{p}{q}$ where p, q ∈ Z and $q \neq 0$.
• Irrational numbers are real numbers that cannot be expressed as a ratio of integers (e.g., $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\pi \approx 3.14159$, $<br /> e \approx 2.71828$).

Real Number Line and Ordering on R

• The real number system can be visualized as points on a line with positive direction to the right and 0 as the origin.
• Real numbers are ordered: for any two real numbers a and b, exactly one of the following holds: a > b, a < b, or a = b.
• Symbols '
• Geometrically, a < b means a lies to the left of b on the number line.
• $a ≤ b$ means either a < b or $a = b$.

Definition: Intervals

• An interval is a set of real numbers representing a segment of the real number line.
• Open Interval: For a < b , the open interval from a to b is denoted as (a, b) = {x ∈ R | a < x < b} (endpoints not included).
• Closed Interval: For a < b , the closed interval from a to b is denoted as $[a, b] = {x ∈ R | a ≤ x ≤ b}$ (endpoints included).
• Half-Open Intervals: [a, b) = {x ∈ R | a ≤ x < b} and (a, b] = {x ∈ R | a < x ≤ b} .
• Infinite Intervals: $(-∞, a] = {x ∈ R | x ≤ a}$, (-∞, a) = {x ∈ R | x < a} , $[a, ∞) = {x ∈ R | a ≤ x}$, (a, ∞) = {x ∈ R | a < x} .
• $R = (-∞, ∞)$.
• $±∞$ can never be included in an interval.

Absolute Value

• Definition: $|x| = \begin{cases} x, &amp; \text{if } x ≥ 0 \ -x, &amp; \text{if } x &lt; 0 \end{cases}$

Examples

• $|3| = 3$
• $|-4| = 4$
• $|1 - π| = π - 1$

Properties of Absolute Value

• (i) $|ab| = |a| |b|$
• (ii) $|a|^2 = a^2 \implies |a| = \sqrt{a^2}$ (positive square root)
• (iii) $|a^n| = |a|^n$, $n ∈ Z$, with a ≠ 0 for nonpositive n.
• (iv) $|a + b| ≤ |a| + |b|$ (triangle inequality)

Convention for $\sqrt{}$

• For a > 0 , $\sqrt{a}$ always denotes the positive solution of $x^2 = a$.
• Thus, $\sqrt{4} = 2$.
• Solutions to $x^2 = a$ for a > 0 are $x = ±\sqrt{a}$.

Complex Numbers

• Introduced in the 16th century to obtain roots of polynomial equations.
• A complex number is of the form $z = x + iy$ where $x, y ∈ R$ and i is a symbol satisfying $i^2 = -1$.
• x is the real part of z, and y is the imaginary part of z.
• The set of all complex numbers is denoted C.
• Example: $3 - 2i ∈ C$.

Example

• The real part of $3 - 2i$ is 3, and the imaginary part is -2 (not $-2i$).
• Complex numbers can be added and multiplied by replacing $i^2$ with -1.
• Example: $(2i)^2 = 4i^2 = -4$.

Example: Simplify (3 − 2i)(1 + i)

$(3 - 2i)(1 + i) = 3 + 3i - 2i - 2i^2 = 3 + i + 2 = 5 + i$

Example: Suppose a, b ∈ R. Simplify (a + bi)(a − bi)

$(a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2$

• Complex Conjugate: If $z = a + bi$ is a complex number, the complex conjugate of z is $a - bi$, denoted as $\overline{z}$.
• Example: The complex conjugate of $3 + 2i$ is $\overline{3 + 2i} = 3 - 2i$.
• $z \overline{z}$ is always a real number.

Example: Simplify $\frac{3 - 2i}{1 - i}$

• Multiply top and bottom by the complex conjugate of the denominator:

$\frac{3 - 2i}{1 - i} = \frac{3 - 2i}{1 - i} × \frac{1 + i}{1 + i} = \frac{(3 - 2i)(1 + i)}{1^2 + 1^2} = \frac{1}{2}(3 - 2i)(1 + i) = \frac{1}{2}(5 + i) = \frac{5}{2} + \frac{1}{2}i$

• A polynomial of degree n always has n complex roots (counted with their correct multiplicity).
• Every quadratic has two roots.

Example: Find the roots of $x^2 + 2x + 2 = 0$

• Using the quadratic formula:

$x = \frac{1}{2}(-2 ± \sqrt{4 - 8}) = \frac{1}{2}(-2 ± 2i) = -1 ± i$

• Completing the square:

$x^2 + 2x + 2 = (x + 1)^2 + 1 = 0 \iff (x + 1)^2 = -1 = i^2 \iff x + 1 = ±i$

$\therefore x = -1 ± i$

Polar Form

• Real numbers are represented on the real line.
• A complex number $z = x + iy$ is represented by a point in the complex plane (horizontal axis = real axis, vertical axis = imaginary axis).
• z can be specified by length r (modulus) and angle θ (argument).
• r is the modulus of z, denoted as $|z|$, measuring the distance of z from the origin.
• θ is the argument of z.
• Relationships:

$x = r \cos θ$
$y = r \sin θ$
$z = x + iy = r(\cos θ + i \sin θ)$
$r = |z| = \sqrt{x^2 + y^2}$
$\tan θ = \frac{y}{x} \text{ if } x \neq 0$

Example: Write z = 1 + i in polar form

• Modulus: $|z| = \sqrt{1 + 1} = \sqrt{2}$.
• Argument: $\theta = \frac{\pi}{4}$.

Euler’s Formula

• Euler’s formula states that for any real number θ:

$\cos θ + i \sin θ = e^{iθ}$

• Every complex number $z = x + iy$ can be represented in polar form as $z = re^{iθ}$. This requires defining the exponential function for complex arguments.

Functions

Definition: Function, Domain, Range

• Let X and Y be subsets of R.
• A function $f : X → Y$ is a rule assigning each element $x ∈ X$ to exactly one element $f(x) ∈ Y$, called the value of f at x.
• X is the domain of f.
• $f(X) = {f(x) | x ∈ X}$ is the range of f, also written range(f).
• The range of f, f(X), is a subset of Y.
• The range is the set of all possible values of $f(x)$ as x varies throughout the domain.
• f(X) is not necessarily equal to all of Y.

Example

• The function $f : R → R$ such that $f(x) = x^2$ is a function with domain R.
• The range is given by:

$f(R) = {f(x) | x ∈ R} = {x^2 | x ∈ R} = {y ∈ R | y ≥ 0} = [0, ∞) ⊆ R$

Example

g : (-6, 7) → R
g(x) = 
  {-5,   -6 < x < 0 
  -π,   x = 0
  x,    0 < x < 7

• Domain of g is the open interval $(-6, 7)$.
• Range: $g((-6, 7)) = {-5} ∪ {-π} ∪ (0, 7)$.

Graphs

• A function can be represented by drawing its graph, which is the set of all points $(x, y)$ in a plane where $y = f(x)$.

Convention (Domain)

• An expression like "the function $y = \sqrt{1 - x^2}$" means the function f with $y = f(x) = \sqrt{1 - x^2}$.
• When the domain is not specified, it is taken to be the largest subset of R on which the rule is defined (and gives a real output).
• For $y = \sqrt{1 - x^2}$, the domain would be $[-1, 1]$.

Vertical Line Test

• Not every curve represents the graph of a function.
• For each x in the domain, there must correspond exactly one y in the range.
• In the graph of a function, any vertical line $x = constant$ must cut the graph in at most one point.
• The equation of a circle $x^2 + y^2 = 1$ does not give rise to a function on any domain intersecting $(-1, 1)$.

Example

• Both $y = \sqrt{1 - x^2}$ (the top half of the circle) and $y = -\sqrt{1 - x^2}$ (the bottom half of the circle) are functions of x.
• $f(x) = \sqrt{1 - x^2}; g(x) = -\sqrt{1 - x^2}$.
• $dom(f) = dom(g) = [-1, 1]$, $range(f) = [0,1]; range(g) = [-1, 0]$.

Exponential Functions

• An exponential function is of the form $f(x) = a^x$, where the base a is a positive constant, and x is the exponent or power.
• One common exponential function is $f(x) = e^x$.
• Exponential functions are useful for modeling population growth (base a > 1) and radioactive decay (base 0 < a < 1).

Example

• The half-life of strontium-90, $^{90}Sr$, is 29 years, meaning half of any quantity will disintegrate in 29 years.
• If the initial mass is 24mg, the expression for the mass remaining after t years is:

Mass remaining at time t: $m(t)$
Initial mass: $m(0) = m<em>0$ Half-life: $m(h) = \frac{m</em>0}{2}$
So, $m(2h) = \frac{m<em>0}{4}, m(3h) = \frac{m</em>0}{8}, …, m(nh) = \frac{m_0}{2^n}$

Thus, $m(xh) = \frac{m<em>0}{2^x} = m</em>0 \cdot 2^{-x}$, for $x ∈ R, x ≥ 0$.

$m(t) = m(\frac{t}{h}h) = \frac{m<em>0}{2^{t/h}} = m</em>0 \cdot 2^{-t/h}$

For $m_0 = 24$mg, $h = 29$ years, $m(t) = 24 \cdot 2^{-t/29} mg$.

Composition of Functions

• Let f and g be two functions.
• The composition of f and g, denoted $f ◦ g$, is defined by $(f ◦ g)(x) = f(g(x))$.
• $dom(f ◦ g) = {x ∈ dom(g) | g(x) ∈ dom(f)}$

Example

• $f(x) = x^2 + 1, g(x) = \frac{1}{x}$
• Their compositions are given by:

$(f ◦ g)(x) = f(g(x)) = f(\frac{1}{x}) = \frac{1}{x^2} + 1, x \neq 0$
$(g ◦ f)(x) = g(f(x)) = g(x^2 + 1) = \frac{1}{x^2 + 1}, x ∈ R$

• In general, $f ◦ g \neq g ◦ f$.

One-to-One (1-1) Functions

• A function $f : X → Y$ is said to be one-to-one (1-1) or injective if, for all $x<em>1, x</em>2 ∈ X$, $f(x<em>1) = f(x</em>2) ⇒ x<em>1 = x</em>2$.
• Equivalently: $x<em>1 \neq x</em>2 ⇒ f(x<em>1) \neq f(x</em>2)$.
• On the graph of f, any horizontal line $y = constant$ cuts through the curve in at most one place.

Example

• Show that the function f defined by $f(x) = \sqrt[3]{2x - 5}$ is 1-1.
• The domain of f is R.
• Need to show that $f(x<em>1) = f(x</em>2) ⇒ x<em>1 = x</em>2$ for any $x<em>1, x</em>2 ∈ R$.
• $f(x<em>1) = f(x</em>2) ⇒ \sqrt[3]{2x<em>1 - 5} = \sqrt[3]{2x</em>2 - 5} ⇒ 2x<em>1 - 5 = 2x</em>2 - 5 ⇒ 2x<em>1 = 2x</em>2 ⇒ x<em>1 = x</em>2$
• Therefore, f is 1-1.

Inverse Functions

• Let $f : X → Y$ be a 1-1 function.
• For each $y ∈ f(X)$ (the range of f), there is a unique x with $f(x) = y$.
• Define the inverse function $f^{-1} : f(X) → X$ by $f^{-1}(y) = \text{that unique } x ∈ X \text{ with } f(x) = y$.
• So $f^{-1}(y) = x \iff y = f(x)$.
• The inverse function reverses the direction of the mapping: $f : x ↦ y$ but $f^{-1} : y ↦ x$.

$f^{-1}(y) = x \text{ and } y = f(x) \text{ so } f^{-1}(f(x)) = x \text{ for all } x ∈ X$

$f(f^{-1}(y)) = y \text{ for all } y ∈ f(X)$

• $dom f^{-1} = range f$
• $range f^{-1} = dom f$
• f must be 1-1 in order that $f^{-1}$ be a function.

How to Find \f^{-1}

• $f(x) = y ⇒ f^{-1}(y) = x$
• To find $f^{-1}$, solve for x in terms of y.

Example

• Find $f^{-1}$ if $f : R → R, f(x) = \sqrt[3]{2x - 5}$
• f is 1-1.
• $dom(f) = R$ and $f(R) = R$
• $f^{-1} : f(R) → dom(f)$, so $f^{-1} : R → R$.

$y = f(x) = \sqrt[3]{2x - 5}$

Solve for x in terms of y:

$y^3 = 2x - 5$

$2x = y^3 + 5$

$x = \frac{1}{2}(y^3 + 5)$

So, $f^{-1}(y) = \frac{1}{2}(y^3 + 5)$.

The name of the variable is irrelevant, so we can write $f^{-1}(x) = \frac{1}{2}(x^3 + 5)$.

To obtain the graph of $f^{-1}(x)$, we reflect the graph of $f(x)$ about the line $y = x$.

Example

• $f : R → R, f(x) = x^2$ is not 1-1 and therefore has no inverse.
• However, $x ≥ 0$ gives a 1-1 function $f : [0, ∞) → R, f(x) = x^2$ with range $[0, ∞)$.
• The inverse of this function is then $f^{-1} : [0, ∞) → [0, ∞), f^{-1}(x) = \sqrt{x}$.
• Similarly, the negative half of the function $f(x) = x^2$ is 1-1, with inverse $f^{-1} : [0, ∞) → (-∞, 0], f^{-1}(x) = -\sqrt{x}$.
• This technique is often used when the function is not 1-1 over its entire domain: just take a part where it is 1-1 and determine the inverse for that part.

Logarithms

• Logarithms are the inverse functions of the exponential functions.
• From the graph of $y = a^x$ (a ≠ 1, a positive constant), we see that it is 1-1 and thus has an inverse, denoted $\log_a x$.
• From this definition we have the following facts:

$\log<em>a(a^x) = x \quad \forall x ∈ R$ a^{\loga x} = x \quad \forall x > 0

Domain of $\log_a(x)$ is $(0, ∞)$ and its range is R.

Natural Logarithm

• Now we set $a = e$ (Euler’s number = 2.71828 . . .).
• The inverse function of $f(x) = e^x$ is $\log_e(x) ≡ \ln x$.

Properties

• Using exponent laws, together with the fact that $\ln x$ is the inverse function of $e^x$, we can prove the following. For x, y > 0, we have:

1. $\ln(xy) = \ln(x) + \ln(y)$;
2. $\ln(\frac{x}{y}) = \ln(x) - \ln(y)$; and
3. $\ln(x^r) = r \ln x, r ∈ R$.
4. $\ln(1) = 0$
5. $\ln(e) = 1$

Note $\ln(\frac{1}{x}) = \ln(x^{-1}) = -\ln x$.

Proof property (1)

• Let $\ln(x) = a$. By definition of inverse function this means $e^a = x$.
• Let $\ln(y) = b$, so $e^b = y$.

It follows that $\ln(xy) = \ln(e^a e^b) = \ln(e^{a+b}) = a + b = \ln(x) + \ln(y)$.

Thus $\ln(xy) = \ln(x) + \ln(y)$.

The other properties are proved similarly.

Example: Bacteria Population

• If a bacteria population starts with 100 bacteria and doubles every 3 hours, then the number of bacteria n after t hours is given by the formula $n = f(t) = 100 \cdot 2^{t/3}$.

(a) Find the inverse of this function and explain its meaning.

(b) When will the population reach 50000?

Solution:

a

If $n = 100 \cdot 2^{t/3}$, then

$\frac{n}{100} = 2^{t/3}$

Taking the natural logarithm of both sides gives

$\ln(\frac{n}{100}) = \ln(2^{t/3}) = \frac{t}{3} \ln(2)$

So, $t = \frac{3 \cdot \ln(n/100)}{\ln 2}$

The inverse function describes how long it will take to obtain n bacteria.

b

Substituting $n = 50,000$ gives $t ≈ 26.9$ hours.

Inverse Trigonometric Functions

• The function $y = \sin x$ is 1-1 if we just define it over the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
• The inverse function for this part of $\sin x$ is denoted $\arcsin x$.
• $\arcsin x$ is defined on the interval $[-1, 1]$ and takes values in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
• Similarly, $y = \cos x$ is 1-1 on the interval $[0, \pi]$ and its inverse function is denoted $\arccos x$.
• The function $\arccos x$ is defined on $[-1, 1]$ and takes values in the range $[0, \pi]$.
• Also, $\tan x$ is 1-1 on the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ with inverse function denoted by $\arctan x$.
• Hence $\arctan$ has the domain $(-\infty, \infty)$ with values in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$.