Learning math: relations and functions

These are my solutions to the problems in the book Maths: A Student's Survival Guide.

Direct proportion

A.k.a. "suoraan verrannollisuus".

Remember: $\propto$ means "is proportional to". Another symbol that means the same thing is $\sim$.

Direct proportion formula: $y = kx$, where $k$ is a constant.

$A = \pi r^2 \qquad \text{Area of a circle} \\ A \propto r^2 \qquad \text{When }r^2\text{ doubles, }A\text{ also doubles}$

We can also say that $\dfrac{A_1}{r^2_1} = \dfrac{A_2}{r^2_2} = \ldots$

$f(x) = kx \\ k = \pi \\ x = r^2$

set xrange [-1:5]
set yrange [-1:10]
set xtics 1
set ytics 1
set ylabel "A"
set xlabel "r^2"

f(x) = pi * x

plot f(x) title "f(r^2)"

Note: if you mistakenly graph $A \propto r$, you'd get this:

set xrange [-1:5]
set yrange [-1:15]
set xtics 1
set ytics 1
set ylabel "A"
set xlabel "r"

f(x) = pi * x * x

plot f(x) title "f(r)"

This is wrong, because $A$ is not proportional to $r$, but rather to $r^2$. When $r^2$ doubles, we also expect $A$ to double, because they're directly proportional.

3.A.(b), example 2

$V = \pi r^2 h \qquad \text{Volume of a cylinder} \\ V \propto h \\ \frac{V_1}{h_1} = \frac{V_2}{h_2} \\ V = kh \\ k = \pi r^2$

3.A.(c), example 3

$V = \frac{4}{3} \pi r^3 \qquad \text{Volume of a sphere} \\ V \propto r^3 \\ k = \frac{4}{3} \pi$

3.A.(c), example 4

Period, $T$, of a simple pendulum of length $l$:

$T = 2 \pi \sqrt{l/g} = 2 \pi \frac{\sqrt{l}}{\sqrt{g}} \\ T \propto \sqrt{l} \\ k = 2 \pi \frac{1}{\sqrt{g}} = \frac{2 \pi}{\sqrt{g}}$

3.A.1

(1)

(a):

$V \propto h$

$\dfrac{V_A}{h_A} = \dfrac{V_B}{h_B}$

$\dfrac{V_A}{4} = \dfrac{V_B}{1}$

$\dfrac{V_A}{V_B} = \dfrac{4}{1}$

(b):

$V \propto r$

$\dfrac{V_C}{r^2_C} = \dfrac{V_D}{r^2_D}$

$\dfrac{V_C}{1} = \dfrac{V_D}{16}$

$\dfrac{V_C}{V_D} = \dfrac{1}{16}$

(2)

$E = \frac{1}{2}M v^2 \\ E \propto v^2$

$\dfrac{E_1}{v^2_1} = \dfrac{E_2}{v^2_2}$

$\dfrac{E_1}{25} = \dfrac{E_2}{900}$

$\dfrac{E_2}{E_1} = \dfrac{900}{25} = 36$

(3)

$V = \frac{4}{3} \pi r^3 \\ V \propto r^3$

$\dfrac{V_1}{r^3_1} = \dfrac{V_2}{r^3_2}$

$\dfrac{V_1}{8} = \dfrac{V_2}{512}$

$\dfrac{V_1}{V_2} = \dfrac{8}{512} = \dfrac{1}{64}$

(4)

$T = 2 \pi \sqrt{l/g} \\ T \propto \sqrt{l}$

$\dfrac{T_1}{\sqrt{l_1}} = \dfrac{T_2}{\sqrt{l_2}}$

$\dfrac{T_1}{3} = \dfrac{T_2}{5}$

$\dfrac{T_1}{T_2} = \dfrac{3}{5}$

Partial direct proportion

Remember: the difference between direct and partial direct proportion is basically this:

• Direct proportion: $y = mx$
  • This line goes through the origin of the graph
• Partial direct proportion: $y = mx + c$
  • This line has a y-intercept of something else than 0

For example, if you have a car whose maintenance and other costs are 4000 per year, and you bought the car for 20 000, the cumulative cost at year $x$ is:

$f(x) = 4000x + 20000$

set xrange [0:10]
set yrange [0:80000]
set xtics 1
set ytics 10000
set ylabel "Cost"
set xlabel "Year"

f(x) = 4000 * x + 20000

plot f(x)

Inverse proportion

A.k.a. "kääntäen verrannollisuus".

$xy = k \qquad \text{ or } \qquad y = \frac{k}{x}$

$x \propto \frac{1}{y}$

When $x$ gets larger, $y$ gets proportionally smaller, and vice versa.

If $x$ gets 2 times larger, $y$ gets 2 times smaller: $2x \frac{1}{2}y = k$

If $x$ gets 10 times smaller, $y$ gets 10 times larger: $\frac{1}{10}x 10y = k$

set xrange [-4:10]
set yrange [-4:10]
set xtics 1
set ytics 1

f(x) = 2 / x

plot f(x) title "2/x"

$x$ and $y$ never reach 0. Therefore both $x$ and $y$ is an asymptote, i.e. a line that a curve approaches, but never reaches.

Example: travel time versus speed for a 100 km distance: $f(x) = \frac{100}{x}$

set xrange [0:80]
set yrange [0:10]
set xtics 10
set ytics 1
set xlabel "Speed (km/h)"
set ylabel "Travel time (h)"

f(x) = 100 / x

plot f(x)

Example: rectangles with constant area: $A = xy$, where $A = 12$

set xrange [0:12]
set yrange [0:12]
set xtics 1
set ytics 1
set style fill solid 0.1

$data1 << EOD
0 0
0 12
1 12
1 0
EOD

$data2 << EOD
0 0
0 6
2 6
2 0
EOD

$data3 << EOD
0 0
0 3
4 3
4 0
EOD

f(x) = 12 / x

plot f(x), "$data1" with filledcurves notitle, "$data2" with filledcurves notitle, "$data3" with filledcurves notitle

Mixed proportion

Example: Newton's law of gravitation: the force of attraction between two masses $m_1$ and $m_2$ is directly proportional to the product of the masses, but inversely proportional to the distance between them squared.

$F = \dfrac{k m_1 m_2}{r^2}$

Remember: when modeling physical phenomena with a mathematical model, make sure that your model fits the physical situation, and in which situations the model stops working as intended:

For example, the extension of a spring can be predicted for a known load but, if the load is too great, the spring deforms and the new length can no longer be found.

Functions

3.B.(a)

$\text{a) } y = 2x + 3 \\ \text{b) } y = x^2 - 2x - 3 \\ \text{c) } y = \frac{1}{x^2 + 4} \\ \text{d) } y = (3x + 1)^{1/2} = \cancel{\sqrt{3x + 1}} = \pm \sqrt{3x + 1}$

(1)

Is there only one $y$ value for a specific $x$?

• a: yes, because this is a straight line
• b: yes, because this is a quadratic
• c: yes. This looks like some kind of a normal distribution.
• d: I think not, because a square root gives two values, e.g. $y = \cancel{\sqrt{4}} = \pm \sqrt{4} = \pm 2$
  • Remember: the convention is that $\sqrt{x}$ only means positive square root, whereas $x^{1/2}$ means both the positive and the negative. With $\sqrt{}$ you need to explicitly say plus-minus: $\pm \sqrt{x}$

(2)

Can two different $x$ give the same $y$ value?

• a: no
  • More generally, for $y = mx + c$, yes, if the slope is 0
• b: yes, because a quadratic gives a symmetric parabola
• c: yes. This looks like some kind of a normal distribution when graphed.
• d: not sure. When graphing this, the curve looks like it always changes y-value for each $x$. It looks like there exists a line that is an asymptote, i.e. a line that the curve is approaching, but never reaches. Therefore I think "no".
  • Correct, but note that Desmos only shows the positive $y$ values for $\sqrt{3x + 1}$ and $(3x + 1)^{1/2}$. The correct way to graph it is to have a quadratic parabola that is rotated 90 degrees. The book demonstrates how to get $y = -5$:
    • $y = (3x + 1)^{1/2}$
    • $-5 = (3x + 1)^{1/2}$
    • $(-5)^2 = ((3x + 1)^{1/2})^2$
    • $25 = 3x + 1$
    • $x = 8$

(3)

Can $x$ be any real number?

• a: yes
• b: yes
• c: yes
• d: I think yes and no. No, if you want the output to be a real number, because e.g. $\sqrt{-2}$ probably requires an imaginary number. Maybe yes, if you allow the output to have imaginary numbers.
  • Correct

(4)

If $x$ is the largest set of all possible input values (e.g. $x$ can be any real or imaginary number), what is the complete set of $y$ values?

Remember: $x$ is the set of input values (i.e. the domain) of the function. $y$ is the set of output values (i.e. the codomain).

• a:
  • $x \in \R$
  • $y \in \R$
    • More generally, for $y = mx + c$, if the slope $\neq 0$, then $y \in \R$, because this is a straight line. If the slope $= 0$, then $y \in \{3\}$, that is, $y = 3$.
• b:
  • Looking at the equation, I can see that I can put in any real number, so therefore $x \in \R$. I was wondering if there exist two vertical asymptotes, though, like this:
    
    But this is probably not the case, since I should be able to put in any real number for $x$.
  • As for $y$, the vertex gives me the minimum $y$ value. Therefore $y$ should not be the full set $\R$, but a subset of it, i.e. $\{y \in \R \mid y \geq k \}$, where $k$ is the vertex's $y$ point.
• c:
  • $x \in \R$ I think, like in b above
  • A graph of this curve shows that both the minimum and maximum of $y$ seem to be restricted: https://www.desmos.com/calculator/3uajczu3kp. When $x$ is very large (i.e. approaches infinity), $y$ seems to approach $0$. Therefore I'd say $\{y \in \R \mid y \leq \frac{1}{4} \text{ and } y \gt 0\}$.
    • The book uses interval notation here, i.e. $(0, \frac{1}{4}]$, which means 0 is not inclusive, but $\frac{1}{4}$ is inclusive
• d:
  • If $x \in \R$, then $\{x \in \R \mid x \geq -\frac{1}{3}\}$, because any smaller $x$ results in $\sqrt{-n}$, which is an imaginary number. But if we allow imaginary number, then I guess $x \in ℂ$.
  • If $x \in \R$, then $\{y \in \R \mid y \geq 0\}$. If we allow imaginary numbers, then I'm not sure what $y$ would be.
    • This is incorrect. If $x \in \R$, then $y \in \R$. This is because $(3x + 1)^{1/2} = \pm \sqrt{3x + 1}$, not $\sqrt{3x + 1}$.

Remember: $\sqrt{x}$ does not imply negative and positive square roots! $\sqrt{x}$ is the principal square root of $x$, and is always a non-negative value. Every positive number has two square roots: $\sqrt{x}$ and $-\sqrt{x}$. Therefore you need to explicitly say $\pm \sqrt{x}$.

To graph $\pm \sqrt{3x + 1}$, we need to graph the positive and negative $y$ values separately:

set xrange [-2:10]
set yrange [-3:8]
set xtics 1
set ytics 1

f(x) = sqrt(3*x + 1)
g(x) = -sqrt(3*x + 1)

plot f(x) title "sqrt(3*x + 1)", g(x) title "-sqrt(3*x + 1)"

Important definitions of functions

A function is a particular form of relationship.

• If the relationship $y = f(x)$ is a function, then, for a specific value of $x$, there is only one value of $y$
  • $y = \sqrt{3x + 1}$ is a function, because $y$ can only have a single value per $x$
  • $y = (3x + 1)^{1/2}$ is not a function, because $y$ can have multiple values per $x$
  • The "raindrop test": a vertical line on a graph must not cut the curve more than once, if the curve is of a function. (See vertical line test.)
• The function $y = f(x)$ is one-to-one if, for each value of $y$, there is only one possible $x$, and vice versa
  • $y = 2x + 3$ is one-to-one
  • $y = x^2 - 2x - 3$ is not one-to-one. I guess you could test this with a "horizontal raindrop test". (See horizontal line test.)
• The domain of the function is the set of numbers for the possible values of $x$
  • For example, $\R$
  • In some cases you will want to restrict the possible input values. For example, if $x$ represents a physical size, then a domain of $x \gt 0$ is often suitable.
• The codomain of the function is the set of numbers for the possible values of $y$
  • For example, $\R$
• The range of the function is the set of numbers for the actual possible output values of $y$
  • https://math.stackexchange.com/questions/3317941/what-is-the-difference-between-codomain-and-range
  • The function $f(x) = 0$ has a codomain of $\R$, but a range of $\{0\}$, since the only actually possible output value is $0$

Transforming functions

Remember: horizontal and vertical shifting is also called translating (e.g. translate() in CSS).

Shift vertically

Add a value $a$ to the function's output:

$y = f(x) + a$

set xrange [-4:8]
set yrange [-2:6]
set xtics 1
set ytics 1

f(x) = x*x - x
g(x) = x*x - x + 1
h(x) = x*x - x - 1

plot f(x) title "f(x)", g(x) title "f(x) + 1", h(x) title "f(x) - 1"

Shift horizontally

Add a value $a$ to the function's input:

$y = f(x + a)$

set xrange [-4:8]
set yrange [-2:6]
set xtics 1
set ytics 1

f(x) = x*x - x
g(x) = (x + 1)*(x + 1) - (x + 1)
h(x) = (x - 1)*(x - 1) - (x - 1)

plot f(x) title "f(x)", g(x) title "f(x + 1)", h(x) title "f(x - 1)"

Scale vertically

Multiply/divide the function's output by $a$:

$y = f(x) * a \\ y = f(x) / a$

The scaling is applied through the origin.

set xrange [-4:8]
set yrange [-2:6]
set xtics 1
set ytics 1

f(x) = x*x - x
g(x) = (x*x - x) * 2
h(x) = (x*x - x) / 2

plot f(x) title "f(x)", g(x) title "f(x) * 2", h(x) title "f(x) / 2"

Scale horizontally

Multiply/divide the function's input by $a$:

$y = f(x * a) \\ y = f(x / a)$

The scaling is applied through the origin.

set xrange [-4:8]
set yrange [-2:6]
set xtics 1
set ytics 1

f(x) = x*x - x
g(x) = (x * 2)*(x * 2) - (x * 2)
h(x) = (x / 2)*(x / 2) - (x / 2)

plot f(x) title "f(x)", g(x) title "f(x * 2)", h(x) title "f(x / 2)"

3.B.1

(1)

• b: $f(x) + 2$
• c: $f(x) - 2$

(2)

• b: $g(x + 2)$
• c: $g(x - 2)$

(3)

• b: $h(x) * 2$
• c: $h(x * 2)$

(4)

• b: $p(x) + 2$
• c: $p(x + 2)$

3.B.2

Functions of functions.

(1)

$f(x) = 3x - 5 \\ g(x) = 2x \\ f(g(x)) = 3(2x) - 5 = 6x - 5 \\ g(f(x)) = 2(3x - 5) = 6x - 10$

(2)

$f(x) = x^2 \\ g(x) = 4 - x \\ f(g(x)) = (4 - x)^2 = x^2 - 8x + 16 \\ g(f(x)) = 4 - x^2$

(3)

$f(x) = \frac{1}{x} \\ g(x) = x - 4 \\ f(g(x)) = \frac{1}{x - 4} \\ g(f(x)) = \frac{1}{x} - 4$

Example 1

$g(x) = 2x^2 + 3 \\ g(g(x)) = 2(2x^2 + 3)^2 + 3 = 2(4x^4 + 12x^2 + 9) + 3 = 8x^4 + 24x^2 + 21 \\ g(g(1)) = 8 + 24 + 21 = 53$

Example 1

$f(x) = \frac{2x - 5}{4x + 1}$

$f(3) = \frac{2(3) - 5}{4(3) + 1} = \frac{1}{13}$

$f(x^2) = \frac{2x^2 - 5}{4x^2 + 1}$

$f(2x + 1) = \frac{2(2x + 1) - 5}{4(2x + 1) + 1} = \frac{4x - 3}{8x + 5}$

$f(f(x)) = \dfrac{2(\frac{2x - 5}{4x + 1}) - 5}{4(\frac{2x - 5}{4x + 1}) + 1} = \dfrac{\frac{2(2x - 5)}{4x + 1} - 5}{\frac{4(2x - 5)}{4x + 1} + 1} = \frac{2(2x - 5) - (4x + 1)5}{4(2x - 5) + (4x + 1)1} = \frac{-16x - 15}{12x - 19}$

Inverse functions

If we know a value of $f(x)$ for a particular function can we work out from this what the original value of $x$ must have been?

So far I understand that:

• A function is a mapping from an input value to a single output value
  • Raindrop test: if the same input (i.e. $x$) gives more than one output (i.e. $y$, e.g. $\pm 7$), then it's not a function
  • E.g. Map<x, y> in a programming language
• If a specific output value ($y$) is only achievable with a specific input value ($x$), and no other input value ($x$) gives the same $y$, then the function is called "one-to-one"
  • I.e. a one-to-one mapping from an input value to a unique output value
  • In other words, if $f(a) = f(b)$, then $a = b$. If $a$ and $b$ give the same $y$ value on a graph, then $a$ must be the same as $b$.
  • E.g. Map<y, x> in a programming language

Therefore I'd say that we can find the original value of $x$ if the function is one-to-one. If it's not, then we can't find a single value of $x$ from the function's output.

Remember: a function that undoes the effects of $f(x)$ is called the inverse function of $x$, and is written as $f^{-1}(x)$.

• Only a one-to-one function can have an inverse function
  • A function is only one-to-one if $f(a) = f(b)$ and $a = b$
• $f^{-1}(f(x)) = f(f^{-1}(x)) = x$
  • $f^{-1}(x)$ undoes $f(x)$
  • $f(x)$ undoes $f^{-1}(x)$

Remember: a self-inverse function is an inverse of itself:

$f(f(x)) = x \\ f^{-1}(x) = f(x)$

Remember: another way to write a function is arrow notation:

• $f : x \mapsto x - 3$
• $f^{-1} : x \mapsto x + 3$

$f(x) = x - 2 \\ f^{-1}(x) = x + 2$

set xrange [-4:10]
set yrange [-3:6]
set xtics 1
set ytics 1

f(x) = x - 2
g(x) = x + 2
s(x) = x

plot f(x) title "f(x)", g(x) title "f^{-1}(x)", s(x) title "Line of symmetry, y = x

$g(x) = 2x \\ g^{-1}(x) = \frac{1}{2}x = \frac{x}{2}$

set xrange [-4:10]
set yrange [-3:6]
set xtics 1
set ytics 1

f(x) = 2 * x
g(x) = x / 2
s(x) = x

plot f(x) title "g(x)", g(x) title "g^{-1}(x)", s(x) title "Line of symmetry, y = x

$p(x) = 6 - x = p^{-1}(x) \qquad \text{Self-inverse function}$

set xrange [-6:12]
set yrange [-3:10]
set xtics 1
set ytics 1

f(x) = 6 - x
s(x) = x

plot f(x) title "p(x) = p^{-1}(x)", s(x) title "Line of symmetry, y = x

$q(x) = 12 / x = q^{-1}(x) \qquad \text{Self-inverse function}$