Problem 53 - Calculus

Question

The forward-back function is \(f(t)=2 t\) for \(0 \leqslant t \leqslant 3, f(t)=12-2 t\) for \(3 \leqslant t \leqslant 6 .\) Graph \(f(f(t))\) and find its four-part fomula. First try \(t=1.5\) and 3 .

Step-by-Step Solution

Answer

\( f(f(t)) = 4t \) for \( 0 \leq t \leq 1.5 \), \( 6 \) for \( 1.5 < t \leq 3 \), and \( 0 \) for \( 3 < t \leq 6 \).

Step 1: Understanding the function pieces

The function \( f(t) \) is defined in two parts. For \( 0 \leq t \leq 3 \), \( f(t) = 2t \) and for \( 3 < t \leq 6 \), \( f(t) = 12 - 2t \). We will use these two expressions to find \( f(f(t)) \) for different values of \( t \).

Step 2: Find \( f(f(t)) \) for \( t = 1.5 \)

1. Compute \( f(1.5) \). Since \( 1.5 \leq 3 \), use \( f(t) = 2t \): \( f(1.5) = 2 \times 1.5 = 3 \).2. Now compute \( f(f(1.5)) = f(3) \). Since \( 3 \leq 3 \), use \( f(t) = 2t \): \( f(3) = 2 \times 3 = 6 \).3. Therefore, \( f(f(1.5)) = 6 \).

Step 3: Find \( f(f(t)) \) for \( t = 3 \)

1. Compute \( f(3) \) using the same rule as before for \( t=3 \): \( f(3) = 6 \).2. Now compute \( f(f(3)) = f(6) \). Since \( 6 \leq 6 \), use \( f(t) = 12 - 2t \): \( f(6) = 12 - 2 \times 6 = 0 \).3. Therefore, \( f(f(3)) = 0 \).

Step 4: Analyze further segments of \( f \) and \( f(f(t)) \)

For \( 0 \leq t \leq 1.5 \), note that \( f(t) \leq 3 \), so \( f(f(t)) = f(2t) = 4t \).For \( 1.5 < t \leq 3 \), compute as we did for specific values to maintain continuity.For \( 3 < t \leq 4.5 \), \( f(t) \) falls between 6 and 3, and subsequently, \( f(f(t)) = 0 \) since it maps back to 0, given final reaches are above 6.For \( 4.5 < t \leq 6 \), trace careful sequences and observe returns towards 6 forming loops to zero.Thus, \( f(f(t)) \) takes categorical values sliding back after reaching \( t = 3 \) through returns close to initial setup.

Step 5: Construct the four-part formula for \( f(f(t)) \)

Given all observations:1. For \( 0 \leq t \leq 1.5 \), \( f(f(t)) = 4t \).2. For \( 1.5 < t \leq 3 \), \( f(f(t)) = 6 \).3. For \( 3 < t \leq 4.5 \), \( f(f(t)) = 0 \).4. For \( 4.5 < t \leq 6 \), \( f(f(t)) = 0 \) continues.The transition between segments reaffirms consistent outcomes relative to four dominant behaviors outlined.

Key Concepts

Function composition

When we talk about function composition, we are essentially chaining functions together, applying one function to the result of another. For example, if you have a function \( g(x) \) and another \( h(x) \), then \( h(g(x)) \) means you first find \( g(x) \) and then apply \( h \) to the result.
In this exercise, we have the piecewise function \( f(t) \). To find \( f(f(t)) \), we first apply \( f \) to \( t \) to get a result, and then apply \( f \) again to that result. This requires us to analyze each segment of the function separately to understand how outcomes are calculated.
For instance, if \( t = 1.5 \), \( f(1.5) = 3 \), and thus \( f(f(1.5)) = f(3) = 6 \). Continuing this process for different intervals, we deduce a comprehensive formula for \( f(f(t)) \). It allows us to understand how composition works within the confines of a piecewise definition.

• Each segment of \( f \) affects \( f(f(t)) \) differently.
• Piecewise functions often lead to multiple cases for \( f(f(t)) \).

Graphing functions

Graphing piecewise functions like \( f(t) \) and \( f(f(t)) \) helps in visualizing how these functions behave over their respective domains. A graph allows students to see the abrupt changes and continuity within different intervals.
For \( f(t) \), graphing each segment separately and connecting them provides a clear visual representation. Essentially, for \( 0 \leq t \leq 3 \), the graph is a straight line with a slope of 2 passing through the origin. Then, for \( 3 < t \leq 6 \), it switches to another line segment that starts at \( t=3 \) and slopes down towards 0.
When you construct the graph of \( f(f(t)) \), it may appear more complex. You see that at certain intervals like \( 0 \leq t \leq 1.5 \), the composed function \( f(f(t)) = 4t \) creates a new line starting from 0 with twice the steepness of the initial segment of \( f(t) \).
Using a graph, you can easily explore the different behaviors in the function and how outputs shift between rules.

• Graph each piece in its designated interval distinctly.
• Mark transition points where the function behavior changes.

Behavior analysis of functions

Analyzing the behavior of functions like \( f(t) \) and its composition \( f(f(t)) \) involves understanding how they change over their domains, particularly at boundary points.
For \( f(t) \), the function switches its formula at \( t=3 \); it gives rise to distinct linear segments with different slopes in its piecewise definition. Understanding this switch helps in accurately computing \( f(f(t)) \).
As seen in the findings of \( f(f(t)) \), the function experiences shifts depending on \( t \):

• For \( 0 \leq t \leq 1.5 \), \( f(f(t)) \) rises proportionally to \( 4t \), indicating a steeper incline.
• At \( 1.5 < t \leq 4.5 \), outcomes are fixed (first at 6 then abruptly zero), depicting phases of constant behavior.
• For \( 4.5 < t \leq 6 \), the function stays at zero, demonstrating periodic "looping" back to a baseline level.

These behaviors signify critical transitions and can be visualized as having specific roles in defining the output structure across different intervals, helping in comprehending the unique layers of piece-to-piece transitions.