Tutorial Week 2

Welcome to your first tutorial. In this tutorial, we’ll be reviewing some prerequisite material you’re already expected to understand as well as week 1 material.

Writing Math

Assigning values to variables

You may be familiar with variables like \(x\) or \(y\) from high school, but on something like a test or written assignment, how would you define a new variable?

One way would be to use the word “let”, such as:

\(\text{Let } x = 7\)

\(\text{Let } f(x) = x^2 + 6\)

\(\text{Let } g(t) = t^{84} + 65.99 \text{ represent the rising cost of internet.}\)

Of course, there are many other ways to define a variable, and you might notice some of them while doing your homework/readings.

Showing your work

On a test, try to adequetely justify each step in your solution, as simply stating the answer to a complicated problem may not earn you a lot of marks (or if any at all).

For example, if you were given the question:

What are the roots of \(f(x) = 4x + 1\)?

Your solution could be something like:

\(f(x) \text{ has a root when } f(x) = 0\)

\(\text{So let } f(x) = 0\)

\(\text{so that we have }\)

\[\begin{split}\begin{aligned} 0 &= 4x + 1 \\ -4x &= 1 \\ x &= \dfrac{1}{-4} \end{aligned}\end{split}\]

\(\text{Therefore } f(x) \text{ has a root at } x = \dfrac{1}{-4} \text{.}\)

Factoring

It probably has been a few months since the last time you factored something, so I thought it’ll be good that we review this.

One thing that should sound familiar is the term “perfect square”.

A perfect square is an integer that is the square (\(x^2\)) of another integer.

For example, the perfect square \(9 = 3^2\).

Common factors

If two terms have a common factor, then you can “move” that factor out, for example \(3x + 3y = 3(x + y)\) where \(3\) is the common factor of \(x\) and \(y\).

Difference of squares

If you had an equation in the form \(x^2 - y^2\), then you can factor it into \((x - y)(x + y)\). For example, \(x^2 - 9 = (x-3)(x+3)\)

In this case, \(9 = 3^2\), but the second term (\(y\)) doesn’t have to be an integer. For example, you could have \(x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2})\) since \(2 = (\sqrt{2})^2\).

Cross-multiplication method (for quadratics)

You could find the full diagrams and stuff by googling cross-multiplication method, so here’s a somewhat simplified version just to get the essentials down:

Say you have the quadratic \(ax^2 + bx + c\) and let the factored form be \((gx + j)(hx + k)\). Then you need to find the values of \(g, j, h, k\) such that:

\[\begin{split}\begin{aligned} a &= gh \text{ (The numbers preceding x in the factored form)} \\ c &= jk \text{ (The numbers being subtracted in the factored form)} \\ b &= gk + jh \text{ (Cross multiply)} \end{aligned}\end{split}\]

Quadratic equation

If you still can’t manage to factor your quadratic equation using the previous methods, then you can get the factors by solving the quadratic equation:

\(x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}\)

Lines

Q1: What is the equation of the line with a x-intercept of \(-3\) and a y-intercept of \(5\)?

Show Solution

From what you know about x and y intercepts, you should be able to identify that \((-3, 0)\) and \((0, 5)\) are points on the line. You can find the slope using the slope formula

\[\begin{split}\begin{aligned} m &= \dfrac{y_2 - y_1}{x_2 - x_1} \\ &= \dfrac{0 - 5}{ -(-3) - 0} \\ &= -\dfrac{5}{3} \end{aligned}\end{split}\]

You can plug this into the slope-intercept form of a line, that is \(y = mx + b\) which gives you \(y = -\dfrac{5}{3}x + 5\) (recall how \(5\) is the y-intercept?).

Q2: Find the equation of a line perpendicular to \(-3x + 2y + 4 = 0\).

Show Solution

First, you want to turn this into slope-intercept form, so move everything but \(y\) to the right side to get \(y = \dfrac{3}{2}x - \dfrac{4}{2} = \dfrac{3}{2}x - 2\).

This gives you the slope of the line, which is \(\dfrac{3}{2}\). The slope perpendicular to this would be the negative reciprocol (that is, \(m_2 = -\dfrac{1}{m_1}\)). So you take the negative reciprocol of \(\dfrac{3}{2}\) to get \(m_2 = -\dfrac{2}{3}\).

So any line with the slope \(-\dfrac{2}{3}\) is perpendicular, a simple example being \(y = -\dfrac{2}{3}x\)

Functions

Q3: Suppose you have the function \(f(4x + 1) = \dfrac{3x}{2x + 1}\).

1. What is \(f(5)\)?

Show Solution

First of all, this means that you want to find the value of \(f(5)\), but the only information you have about the function is \(f(4x + 1) = \dfrac{3x}{2x + 1}\). So how can you get \(f(5)\) in there? You would make \(f(4x + 1) = f(5)\), which means you need to find the value of \(x\) such that \(4x + 1 = 5\).

Solving this equation gets us

\[\begin{split}\begin{aligned} 4x + 1 &= 5 \\ 4x &= 5 - 1 \\ 4x &= 4 \\ x &= 1 \end{aligned}\end{split}\]

So to get \(f(5)\), you need \(x = 1\). You can now just subtitute \(x = 1\) into \(\dfrac{3x}{2x + 1}\) which gets you

\[\begin{split}\begin{aligned} \dfrac{3x}{2x + 1} &= \dfrac{3(1)}{2(1) + 1} \\ &= \dfrac{3}{3} \\ &= 1 \end{aligned}\end{split}\]

2. What is \(f(x)\) for an arbitrary x?

Show Solution

For this, like the first question, you want to find a value of x that satisfies every value put into \(f(x)\). To avoid having a multiple \(x\) variables in your answer, lets reword it so that we’re looking for the value of \(f(t)\) for an arbitrary \(t\).

You need a value of \(x\) for every \(t\), so let \(t = 4x + 1\).

Solving for \(x\) gives you

\[\begin{split}\begin{aligned} 4x + 1 &= t \\ 4x &= t - 1 \\ x &= \dfrac{t - 1}{4} \end{aligned}\end{split}\]

Now that you have \(x\) in terms of \(t\), you can subtitute your new value of \(x\) into \(\dfrac{3x}{2x + 1}\) which gives you \(f(t) = \dfrac{3(\dfrac{t-1}{4})}{2(\dfrac{t - 1}{4}) + 1} = \dfrac{3t-3}{2(t+1)}\)

With this, you can say that \(f(x) = \dfrac{3x-3}{2(x+1)}\)

3. What is the natural domain of \(f(x)\)?

Show Solution

There is an aymptote when \(2(x+1) = 0\), which has the solution

\[\begin{split}\begin{aligned} \dfrac{2(x+1)}{2} &= \dfrac{0}{2} \\ x + 1 &= 0 \\ x &= -1 \end{aligned}\end{split}\]

This means that the function is defined on all points but \(x = -1\), so the natural domain is \((-\infty , -1) \cup (-1, \infty )\).

Q4: What is the natural domain of \(f(x) = \sqrt{cos(sin(x))}\)?

Show Solution

Since there’s a square root, you know that the domain is all values where \(cos(sin(x)) \geq 0\).

The range of \(sin(x)\) is \([-1, 1]\), but if you look at the graph of \(cos(x)\), it is positive from \([-\dfrac{\pi}{2}, \dfrac{\pi}{2}]\) (\(\dfrac{\pi}{2} = 1.57\)) so \(cos(sin(x))\) is positive for all numbers, which means the domain is \((-\infty, \infty)\).

Q5: Find a function with the domain \([0, 3) \cup (3, 5]\).

Show Solution

For this question, you’ll need to dig into your toolset of functions and find one that has a restriction on “one end” of its domain. The square root function works here, since the value being square rooted can’t be negative.

So, let’s identify the restrictions on the domain:

1. \(x \geq 0\)

2. \(x \leq 5\)

3. \(x \neq 3\)

For the first restriction, you can use the function \(\sqrt{x}\).

For the second restriction, you can use the function \(\sqrt{5 - x}\). (or the square root of any function that is positive on \((-\infty, 5]\))

For the last restriction, divide your function by \(x - 3\), since that will create an asympotote at \(x = 3\).

Now, combining these restriction, we get \(f(x) = \dfrac{\sqrt{x}\sqrt{5 - x}}{x - 3} = \dfrac{\sqrt{x(5-x)}}{x-3}\).

Q6: Find the domain of \(f(x) = |x + 4| - x^2\).

Show Solution

If you have the sum or difference of two functions (in this case \(|x + 4|\) and \(x^2\)), then the domain is the intersection of the two.

Thus, since the domain of both functions in this case is all real numbers, the domain is \(( -\infty, \infty )\).

Sketching functions

Check your answers using https://www.desmos.com/calculator.

Q7: Sketch \(y = 3log_2(x + 4) - 2\).

Show Solution

Let’s first get the graph of \(log_2(x)\).

You know that \(log_2(x)\) is the inverse of \(2^x\), so using the table of values for \(2^x\):

x	y
-2	0.25
-1	0.5
0	1
1	2
2	4

we can switch the \(x\) and \(y\) columns to get the table of values for \(log_2(x)\).

x	y
0.25	-2
0.5	-1
1	0
2	1
4	2

From the equation, you can see that the following transformations were applied on the graph of \(log_2(x)\)

1. Vertical stretch by a factor of 3

2. Vertical shift down 2 units

3. Horizontal shift left 4 units

You can plot these transformations on the table of values and then graph.

\(x - 4\)	\(3y - 2\)
-3.75	-8
-3.5	-5
-3	-2
-2	1
0	4

Q8: What is the domain of \(y = 3log_2(x + 4) - 2\)?

Show Solution

Recall that you can only take the log of a positive number (which means 0 is also excluded), so solving for the inequality \(x + 4 > 0\) gives us \(x > -4\). The domain is \((-4, \infty)\).

Q9: Sketch \(y = \sqrt{4 - x^2}\).

Show Solution

Plot the points, and you’ll see the upper half of a circle with radius 2 centered around the origin.

The equation of a circle is \(x^2 + y^2 = r^2\), where \(r\) is the radius.

So, manipulating our equation gives us

\(y^2 = 4 - x^2\)

\(y^2 + x^2 = 2^2\)

Which is a circle with radius 2. But, note that from the original equation, \(y\) is always positive, so we only take the part of the circle where \(y \geq 0\).

Function inverses

Q10: Find the inverse of \(y = e^x - 7\).

Show Solution

Switch x and y in the equation, and then solve for y. We get:

\[\begin{split}\begin{aligned} x &= e^y - 7 \\ x + 7 &= e^y \\ ln(x + 7) &= ln(e^y) \\ ln(x + 7) &= y \end{aligned}\end{split}\]

So the inverse if \(y = ln(x + 7)\)

Q11: Find the inverse of \(y = x^2\) restricted on \(x \geq 0\).

Show Solution

Without the domain restriction, \(y = x^2\) wouldn’t pass the horizontal line test and therefore wouldn’t have an inverse. But with this domain restriction, we can find such an inverse for the function.

So you can do:

\[\begin{split}\begin{aligned} x &= y^2 \\ \sqrt{x} &= \sqrt{y^2} \\ \sqrt{x} &= |y| \\ y &= \pm \sqrt{x} \end{aligned}\end{split}\]

But note how when you take the inverse of a function, the domain and range switch. Since your domain is \(x \geq 0\), your range for your inverse is \(y \geq 0\) which means that the equation \(y = \pm \sqrt{x}\) should actually be \(y = \sqrt{x}\), and this, is also your inverse.

Q12: Let \(f(x) = e^{x^3 - 1} + 2x - 1\). Evaluate \(f^{-1}(\dfrac{1}{e} - 1)\).

Show Solution

What we are looking for is the value of \(x\) that maps \(f(x)\) to \(\dfrac{1}{e} - 1\).

Make a table of values for f(x).

x	y
-2	\(e^{-9} - 5\)
-1	\(e^{-2} - 5\)
0	\(e^{-1} - 1\)
1	\(2\)
2	\(e^{7} + 4\)

When \(x = 0\), we have what we want! So \(f^{-1}(\dfrac{1}{e} - 1) = 0\).

Absolute values

Q13: Solve the equation \(\dfrac{|x + 4|}{|x - 1|} = 5\).

Show Solution

\[\begin{split}\begin{aligned} \dfrac{|x + 4|}{|x - 1|} &= 5 \\ |\dfrac{x + 4}{x - 1}| &= 5 \\ \dfrac{x + 4}{x - 1} &= \pm 5 \end{aligned}\end{split}\]

Solve for \(\dfrac{x + 4}{x - 1} = 5\) and \(\dfrac{x + 4}{x - 1} = -5\).

Q14: Solve the inequality \(|x - 5| \lt 3\).

Show Solution

Using inequality properties, we can get

\[\begin{split}\begin{aligned} -3 \lt\; &x - 5 \lt 3 \\ -3 + 5 \lt\; &x - 5 + 5 \lt 3 + 5\\ 2 \lt\; &x \lt 8 \end{aligned}\end{split}\]

Inequalities

Q15: Solve the inequality \(x^3 + 3x^2 > -2x\).

Show Solution

Move the \(-2x\) to the left side to get \(x^3 + 3x^2 + 2x > 0\).

Factor out \(x\) to get \(x(x^2 + 3x + 2) > 0\).

Factor the quadratic to get \(x(x+2)(x+1)>0\).

Let \(f(x) = x(x+2)(x+1)\). Now, use a number line, mark the x-intercepts, and check the values in between the x-intercepts to see if they’re greater than 0.

\(f(x)\) is useful here since you can show your calculations by writing something like \(f(-0.5)=-\dfrac{3}{8} < 0\) for your number line.

Logarithms and exponents

Q16: Solve the equation \(2(\pi^x) + 7(\pi^x) -e^8 = 0\).

Show Solution

\[\begin{split}\begin{aligned} 2(\pi^x) + 7(\pi^x) -e^8 &= 0 \\ 9(\pi^x) -e^8 &= 0 \\ 9(\pi^x) &= e^8 \\ \pi^x &= \dfrac{e^8}{9} \\ log_{\pi}(\pi^x) &= log_{\pi}(\dfrac{e^8}{9}) \\ x log_{\pi}(\pi) &= log_{\pi}(\dfrac{e^8}{9}) \\ x (1) &= log_{\pi}(\dfrac{e^8}{9}) \\ x &= log_{\pi}(\dfrac{e^8}{9}) \\ \end{aligned}\end{split}\]

Function composition

Q17: Let \(f(x) = x^2 - x\), \(g(x) = log_{\scriptsize{}\dfrac{1}{2}}(x)\), \(h(x) = cos(x)\). Restrict \(x\) to \(0 \leq x \leq \pi\). Solve the equation \((f \circ g \circ h)(x) = 0\).

Show Solution

You can somewhat simplify this question by breaking it down into multiple steps.

Let’s first find the values of \(x\) that satisfies \(f(x) = 0\).

\[\begin{aligned} x^2 + 10x + 16 &= x(x-1) \end{aligned}\]

The values of \(x\) where \(f(x) = 0\) in this case are \(x = 0\) and \(x = 1\), so you want to find the values of \(t\) (since you already used \(x\) for \(f(x)\), so you want to avoid getting confused with variables with the same name) that makes \(g(t) = 0\) or \(g(t) = 16\).

For \(g(t) = 0\), you have

\[\begin{split}\begin{aligned} g(t) &= 0 \\ log_{\scriptsize{}\dfrac{1}{2}}(t) &= 0 \\ \dfrac{1}{2}^{(log_{\tiny{}\dfrac{1}{2}}(t))} &= (\dfrac{1}{2})^0 \\ t &= 1 \end{aligned}\end{split}\]

For \(g(t) = 8\), you have

\[\begin{split}\begin{aligned} g(t) &= 1 \\ log_{\scriptsize{}\dfrac{1}{2}}(t) &= 1 \\ \dfrac{1}{2}^{(log_{\tiny{}\dfrac{1}{2}}(t))} &= (\dfrac{1}{2})^1 \\ t &= \dfrac{1}{2} \end{aligned}\end{split}\]

So your values of \(t\) are \(t = 1\) and \(t = \dfrac{1}{2}\).

Now you need to find the values of \(u\) (again, using another variable since you already used \(x\) and \(t\)) that makes \(h(u) = 1\) or \(h(u) = \dfrac{1}{2}\).

This means you’re solving for \(cos(u) = 1\) and \(cos(u) = \dfrac{1}{2}\).

Since our domain of \(u\) is restricted on \(0 \leq u \leq \pi\), the only values of \(u\) that are solutions to this are \(u = 0\) and \(u = \dfrac{\pi}{3}\) (use a unit circle). And this is our answer: \(x = 0\) and \(x = \dfrac{\pi}{3}\).