Activity 1.1.1. Making Sense of the Slope of One Point.
The goals of this activity are to see what can happen with sets of nearby slopes and to become comfortable working with functions that have no formula.
(a)
First we will consider the polynomial \(y=2x^3-9x^2+10x-3\text{.}\)
(i)
Calculate the points on this polynomial at \(x=1\) and \(x=2\text{.}\)
(ii)
Find the slope of the line between these two points.
(iii)
Repeat this process to complete Table 1.1.1 (you just calculated the entries for the first row). Note technology is great for this task.
(iv)
Do you notice any pattern in the slopes? If you are not confident you see a pattern you could extend the table (calculate more rows).
(v)
Based on this is there a reasonable answer to the question what is the slope at the one point \(x=1.5\text{?}\)
(b)
Next we will consider the Cantor Function.
(i)
Follow these directions to construct the graph of the Cantor function.
Generally the construction adds a horizontal line segment over the middle third of a horizontal interval that is half way between the previously added line segments to its left and right.
(A)
Construct the axes. The x-axis should be from 0 to 1. If using a ruler, draw a line 243 mm long with 0 on the left and 1 at 243 mm (this will make later steps easier). The y-axis should be labeled 0 to 1. The actual height is not important. To make it easy use the full height of the sheet of paper.
(B)
Find the endpoints of the middle third along the x-axis.
For the first step the thirds are \(0=0/3, 1/3, 2/3, 3/3=1\text{.}\) The middle third is \((1/3,2/3)\text{.}\)
If we are starting with the interval \((0,1/3)\) then the thirds are \(0=\frac{0}{3} \cdot \frac{1}{3}, \frac{1}{3} \cdot \frac{1}{3}=\frac{1}{9}, \frac{2}{3} \cdot \frac{1}{3}=\frac{2}{9}, \frac{3}{3} \cdot \frac{1}{3}=\frac{1}{3}\text{.}\)
(C)
Average the height of the horizontal segments to the left and right. If there is no segment to the left use \(y=0\text{.}\) If there is no segment to the right use \(y=1\text{.}\)
For the first segment the height is \(\frac{1}{2}(0+1)=\frac{1}{2}\text{.}\)
For a segment over the interval \((1/9,2/9)\) the height on the left is 0 (no segment) and the height on the right is \(1/2\) so the height of this segment is \(\frac{1}{2}\left(0+\frac{1}{2}\right)=\frac{1}{4}\text{.}\)
(D)
Draw a horizontal line segment over this middle third at the height you just calculated.
(E)
On each of the intervals (thirds from previous steps) repeat this process starting at Task 1.1.1.b.i.B.
(F)
Stop when you can no longer draw any smaller segments.
(ii)
Calculate slopes.
(A)
Calculate the points on the Cantor function at \(x=\frac{1}{3}\) and \(x=\frac{2}{9}\text{.}\)
(B)
Find the slope of the line between these two points.
(C)
Repeat this process to complete Table 1.1.3 (you just calculated the entries for the first row).
(D)
Do you notice any pattern in the slopes? If you are not confident you see a pattern you could extend the table (calculate more rows).
(E)
Based on this is there a reasonable answer to the question what is the slope at the one point \(x=1/3\text{?}\)
