Simulation Of Practice-fund-final

Simulation of Practice: Concept of Inverse Goals for the lesson The goal for this lesson is students to understand and demonstrate they know the concept invertible functions and how to solve for them by using real-world problems, terminology, and multiple representations including correspondence and covariation. Solution to the Opening Task (a) Draw a picture of this pot 6 in.

Total Volume = 2 gallons

(b) Find the relationship between v, for a volume of water poured into the pot, and h, the height of water in the pot at that volume. Graph this relationship.  Relationship: 𝑣 = 𝜋𝑟 2 ℎ  Since we know that the total height of the pot is 6 inches and the total volume of the pot is 2 gallons then we can find what our radius is. 

1 3

( ) 𝜋

1

=𝑟

2

𝑜𝑟

√(3) = 𝑟 𝜋

1 3

( )

 So our new relationship will be: 𝑣 = 𝜋 ( 𝜋 ) ℎ 1

 But since the 𝜋’s cancel each other out then we are left with: 𝑣 = 3 ℎ Graph 1:

3

(c) If there is 4 gallon of water in the pot, how high is the water level? 3

3

1

 So if there is 4 gallon of water in the pot then 4 = 3 ℎ  Then,

3 4 1 3

3

= ℎ meaning that the height of the water level at 4 𝑖𝑠 2.25 𝑜𝑟

9 4

(d) If the water level in the pot is 3 inches high, how much water is in the pot? 1  So if the water level is 3 inches in the pot then 𝑣 = 3 × 3 meaning that the volume of the pot is 𝑣 = 1𝑔. (e) If the water level is 3.5 inches high, how much water is in the pot? Correct answer 1: 1  So if the water level is 3.5 inches in the pot then 𝑣 = 3 × 3.5 meaning that the 7

volume of the pot is 𝑣 = 6 𝑔. Correct answer 2: 1  So if the water level is 3.5 inches in the pot then 𝑣 = 3 × 3.5 meaning that the volume of the pot is 𝑣 = 1.66667𝑔 Incorrect answer:

1

 Relationship: 𝑣 = 𝜋 3 ℎ

1

 So if the water level is 3.5 inches in the pot then 𝑣 = 𝜋 3 × 3.5 meaning that the volume of the pot is 𝑣 = 3.66519142919𝑔 Immediately following the Opening Task  What I would say after the class has completed the activity: “The opening activity consisted of finding the function used when the height is the input and the volume is the output. However, when finding the inverted function (undoing function) of the pot it would consist of having the volume as the input and the height as the output.”  Notation I would use to refer the function being inverted: Usually the notation I would use to refer to the function being inverted I would use 𝑓 −1 , but since that is what we will teach in the key terminology then I would refer to an inverted function as something that “undoes” the other function. I would demonstrated this using a diagram. (Diagram 1)

Function:

Height

Volume

Inverted Function (Undoing):

Volume

Height

 How I would define the function and its inverse: Function: A special relationship where each input has a single output. Inverse Function: A function that undoes the action of another function.  How it has to do with the opening task: 1 Function in opening task: 𝑣 = 3 ℎ Inverse Function in opening task: 𝑣 = ℎ ∕ (1⁄3)

Key terminology  Definition 1. Given an invertible function f, the inverse of f is the function that maps 𝑦 ↦ 𝑥 whenever 𝑥 ↦ 𝑦 is an assignment of f. The inverse function is denoted 𝑓 −1 .  Definition 2. Given an invertible function f, the inverse of f is the function such that for all x in the domain of f, we have 𝑓 −1 ⋅ 𝑓(𝑥) = 𝑥  Phrases or terminology in the definitions that the students would benefit from discussion to understand includes: maps, 𝑦 ↦ 𝑥, 𝑥 ↦ 𝑦, and composite (𝑓 −1 ⋅ 𝑓(𝑥) = 𝑥). Illustrating concept with multiple representation Problem context: We can observe the height (input) is our x and how our volume (output) is our y. Meaning that the function f maps 𝑥 ↦ 𝑦. Which we were able to observe in our diagram 1. Hence, our inverse function 𝑓 −1 would have the volume be our new input and height as our new output. Meaning that 𝑓 −1 maps 𝑦 ↦ 𝑥. Which we were also able to observe in our diagram 1. Now when we observe definition two, we can use this knowledge to prove it correct: (𝑓 −1 (f(x)) = (𝑓 −1 (f(height)) = (𝑓 −1 (volume)) = Height. Diagram 2:

1

Algebraic notation: Since our function is equal to 𝑣 = 3 ℎ and our 𝑓 −1 equals 𝑣 = ℎ⁄(1⁄3) then 𝑓 −1 ⋅ 𝑓(𝑥) equals

1∕3ℎ 1∕3

meaning that the (1/3)’s cancel each other leaving 𝑓 −1 ⋅ 𝑓(𝑥) = h.

Graph: In the graph we see how the x-axis changes from height to volume and how the y-axis changes from volume to height.

Three Questions: 1. Is there a problem in the real world where we can see inverse functions being applied?  Anticipated Answer: In Temperature. Fahrenheit to Celsius and Celsius to Fahrenheit.  Ideal Response: The function would be when converting dollars to pesos where dollars is your input and pesos is your output, and your inverse function (𝑓 −1 ) would be converting pesos to dollars where pesos is your input and dollars is your output. 2. Why do you think that when plugging in x into a composite of a function and its inverse give you back x? Explain.  Anticipated Answer: Because its undoing.  Ideal Response: Since when we do a composite of a function and its inverse we undo the function. For example: (𝑓 −1 (f(x)) = (𝑓 −1 (f(x)) = (𝑓 −1 (y)) = x. 3. How would you re-word definition 2 using definition 1?  Anticipated Answer: Using the bubble graphs x maps y and y maps to x so in the composite it undoes itself which is x.  Ideal Response: Given an invertible function f (𝑥 ↦ 𝑦), the inverse of f (𝑦 ↦ 𝑥) is the function such that for all x in the domain of f we have (𝑓 −1 ⋅ 𝑓(𝑥 ↦ 𝑦)) = (𝑓 −1 ⋅ 𝑓(𝑦 ↦ 𝑥)) = 𝑥). Mathematical equivalence of definitions Ideal responses:  Example: Since function f is Height ↦ 𝑉𝑜𝑙𝑢𝑚𝑒, and the inverse of f is Volume ↦Height we have (𝑓 −1 (f(height)) = (𝑓 −1 (volume)) = Height. Hence both definitions are the same thing since both are undoing each other and given Height as the output.  General Response: Both definitions are the same because they complement each other by saying that the functions undo each other giving us that x would be both the first input and the final output. Summary  To summarize, what we learned today was that when we have a composition of a function (f(x)), and its inverse function (𝑓 −1 ) that these two will end up undoing on another leaving the end result as input=output. Follow up If we have a pot that has a smaller volume at the bottom then at the top, would this statement be true if we graphed the relationship of height as a function of volume: 𝑓 −1 ⋅ 𝑓(𝑥) = 𝑥. Where the total height is 8in and the total volume is 20g. Pot looks like this: