Class 1

During Class

Welcome!

First Day Challenge: Exchange name and contact information with at least 1 other student in this class during our work today.

Brain Gains

Let \(f(x,y) = 4x^2 -5y\sqrt{x+1}\). Find each of the following values.
1. \(f(3,2)\)
2. \(f(0,-2.1)\)
3. \(f(-2.1,5)\)

Answers

\(f(3,2) = 16\)
\(f(0,-2.1) = 10.5\)
\(f(-2.1,5)\) is undefined

Note: The terms “undefined” and “no solution” have different meanings mathematically. In this case, \(f(-2.1,5)\) is undefined because when we try to evaluate \(f(-2.1,5)\), we see \(f(-2.1,5) = 4(-2.1)^2 - 5(5)\sqrt{-2.1+1} = 17.64 - 25\sqrt{-1.1}\). Since there is no real number such that when you square the number you get -1.1, we see \(\sqrt{-1.1}\) is undefined when we are working with real number. You can assume that we will always be working with real numbers in this class, unless otherwise specified.

Consider the piecewise function \[f(x) = \begin{cases} x & \quad x\geq 0 \\ \\ -x & \quad x<0 \end{cases}\] Find the following values.
1. \(f(-2)\)
2. \(f(-27)\)
3. \(f(0)\)
4. \(f(\sqrt{3})\)

Answers

\(f(-2) = 2\)
\(f(-27) = 27\)
\(f(0) = 0\)
\(f(\sqrt{3}) = \sqrt{3} \approx 1.732051\)

The number \(f(\sqrt{3})\) is an irrational number which means any decimal representation of this number is just an approximation. The exact number is represented by the symbol \(\sqrt{3}\) which represents the positive real number that when you square it you get exactly 3.

Group Discussion

We’ll split into randomized groups every day of class. Jot your name on the board to help us all learn each other’s names.
Then work through the following problems.

Take turns acting as scribe for the group. When you are the scribe, it’s fine to not know exactly how to proceed.
The goal is to ask questions of each other and help each other.
Leave up your work so that you can compare your answers with neighbors.

Let \(f(x) = \frac{2x+4}{x^2}\). Find each of the following values.
1. \(f(-4)\)
\(f(1)\)
\(f(0)\)

Answers

\(f(-4) = -0.25\)
\(f(1) = 6\)
\(f(0)\) is undefined

Note: The number \(f(0)\) is undefined in this case because when we try to evaluate \(f(0)\) we see \(f(0) = \frac{2(0)+4}{0^2} = \frac{4}{0}\). What does it mean to divide the quantity 4 into 0 parts? This operation is not defined. Some of you may have had calculus before and tried to give an answer of infinity. Note that infinity is a concept, not a number.

Another way to think of division is that we are looking for a number \(k\) so that \(k = \frac{n}{d}\). Thus \(k\) multiplied by \(d\) would need to be equal to \(n\), in other words \(k d = n\). If the numerator is 4, \(n=4\), and the denominator is 0, \(d = 0\), this would mean \(0 = 4\) (\(k \cdot 0 = 4\)) for any choice of \(k\), which doesn’t make any sense (we have a contradiction). So we say division by zero is undefined.

Let \(f(x) = 8x^2 - 15\). Find each of the following values.
1. \(f(-2)\)
2. \(f(1)\)

Answers

\(f(-2) = 17\)
\(f(1) = -7\)

Consider the piecewise function \(f(x) = \begin{cases} x^2 & \quad x\geq 0 \\ \\ -x+4 & \quad x<0 \end{cases}.\)
Find the following values.
1. \(f(-2)\)
2. \(f(-27)\)
3. \(f(0)\)
4. \(f(\sqrt{3})\)

Answers

\(f(-2) = 6\)
\(f(-27) = 31\)
\(f(0) = 0\)
\(f(\sqrt{3}) = 3\)

Consider the piecewise function \(f(x) = \begin{cases} x^3 & \quad x < -1 \\ \\ -2 & \quad -1 < x < 4 \\ \\ \sqrt{x} & \quad x \geq 4 \end{cases}.\)
Find the following values.
1. \(f(-2)\)
2. \(f(-0.5)\)
3. \(f(3)\)
4. \(f(0)\)
5. \(f(5.2)\)
6. \(f(-1)\)
7. \(f(4)\)

Answers

\(f(-2) = -8\)
\(f(-0.5) = -2\)
\(f(3) = -2\)
\(f(0) = -2\)
\(f(5.2) \approx 2.280351\). The number \(f(5.2)\) an irrational number which means any decimal representation of this number is just an approximation. The exact number is represented by the symbol \(\sqrt{5.2}\) which represents the positive real number that when you square it you get exactly 5.2.
\(f(-1)\) is undefined. We have no rule that tells us what we the output will be when the input is -1. Generally in situations like this, we assume that -1 is not in the domain of \(f\) rather than that the author just forgot to tell us the rule for the output connected to the input -1.
\(f(4) = 2\)

Let \(f(x) = x^2\). Determine whether or not \(f(a+b) = f(a) + f(b)\). Carefully explain your answer. While an example is not an explanation, sometimes examples can be a helpful part of an explanation.

Possible Explanation

Given \(f(x) = x^2\), we will show \(f(a+b) = f(a) + f(b)\) is not true for all values of \(a\) and \(b\) by providing a pair of number \(a\) and \(b\) such that \((a+b)^2 \neq a^2 + b^2\) (a counter example). Let \(a=1\) and \(b=7\). We see \((a+b)^2 = (1+7)^2 = 8^2 = 64\) and \(a^2 + b^2 = 1^2 + 7^2 = 1 + 49 = 50\). We see \((a+b)^2 = 64 \neq 50 = a^2 + b^2\) or \((a+b)^2 \neq a^2 + b^2\).

We can also show that \(f(a+b) = f(a) + f(b)\) is not a true statement by showing that \(f(a+b) \neq f(a) + f(b)\) in general when \(f(x) = x^2\). We see \(f(a) = a^2\) and \(f(b) = b^2\). Now we compute \(f(a+b) = (a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2 \neq a^2 + b^2 = f(a) + f(b)\) when \(2ab \neq 0\). Thus we see that \(f(a+b) \neq f(a) + f(b)\) unless \(2ab = 0\).

Important take away:

The square of a sum is NOT the sum of the squares, \((a + b)^2 \neq a^2 + b^2\).
The operation of addition must happen before the power. The order of operations is grouping and then exponents.

Construct a plot of as many of the functions above as you have time. Use whatever software package you are comfortable with, helping each other to replicate the plot.

Class Discussion

Mastery Learning - Specifications Grading

What is mastery learning?
What is specifications grading?
Course Introduction and Syllabus.
- Are you in the right class?
- Weekly rhythm, what should I be doing between class?
  1. Attendance & Preparation - #tasklist
  2. Knewton-Alta
  3. Projects

Source: Class.1 on byuimath.com