Problem 38 - Contemporary Precalculus

Question

Fill the blanks in the given table. In each case the values of the functions \(f\) and \(g\) are given by these tables: $$\begin{array}{|c|c|} \hline x & f(x) \\ \hline 1 & 3 \\ \hline 2 & 5 \\\ \hline 3 & 1 \\ \hline 4 & 2 \\ \hline 5 & 3 \\ \hline \end{array}$$ $$\begin{array}{|c|c|} \hline t & g(t) \\ \hline 1 & 5 \\ \hline 2 & 4 \\\ \hline 3 & 4 \\ \hline 4 & 3 \\ \hline 5 & 2 \\ \hline \end{array}$$ $$\begin{array}{|c|c|} \hline t & (g \circ g)(t) \\ \hline 1 & \\ \hline 2 & \\\ \hline 3 & \\ \hline 4 & 4 \\ \hline 5 & \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Answer: (g ∘ g)(1) = 2, (g ∘ g)(2) = 3, (g ∘ g)(3) = 3, (g ∘ g)(4) = 4, and (g ∘ g)(5) = 4.

Step 1: Calculate g(1)

First, we need to find the value of \(g(1)\). We can find this value in the table provided for the function \(g(t)\): $$g(1) = 5$$

Step 2: Calculate (g ∘ g)(1)

Now, we will calculate \((g \circ g)(1)\) by substituting the value we found in step 1 into the function \(g(t)\): $$(g \circ g)(1) = g(g(1)) = g(5)$$ Looking up the value in the table we get \(g(5) = 2\). So, \((g\circ g)(1) = 2\).

Step 3: Repeat for all values of t

We must now repeat steps 1 and 2 for the remaining values of \(t\) from 2 to 5: \((g \circ g)(2) = g(g(2)) = g(4) = 3\) \((g \circ g)(3) = g(g(3)) = g(4) = 3\) \((g \circ g)(4)\) has been already given: \((g \circ g)(4) = 4\) \((g \circ g)(5) = g(g(5)) = g(2) = 4\)

Step 4: Complete the table

Now that we have all the missing values, let's fill in the table for \((g \circ g)(t)\): $$\begin{array}{|c|c|} \hline t & (g \circ g)(t) \\\ \hline 1 & 2 \\\ \hline 2 & 3 \\\ \hline 3 & 3 \\\ \hline 4 & 4 \\\ \hline 5 & 4 \\\ \hline \end{array}$$

Key Concepts

Functions

Functions are like machines that take an input, do something to it, and produce an output. They are often represented as \( f(x) \), where \( x \) is the input, and the output is determined based on some rule or table. The function \( f \) corresponds to specific pairs in a table, showing which input leads to which output. For example, if we have a function \( f: \mathbb{R} \to \mathbb{R} \) given by a table, such as:

• When \( x = 1 \), \( f(x) = 3 \)
• When \( x = 2 \), \( f(x) = 5 \)

This table tells us exactly how the function behaves for particular inputs. Remember, the way functions behave is often dictated by specific rules. In some cases, these rules can be summarized in an algebraic expression, while in others, as seen here, a table explicitly lists the values.

Table of Values

Tables of values are a simple way to organize and present data associated with functions. They show exactly what output corresponds to each given input. Think of them like a roadmap telling us the direction from input to output.Typically, tables will have the following format:

• One column for the input values \( (x) \)
• Another column for the output values \( (f(x)) \)

For instance, in our function \( f \) example:

• Input 1 produces an output of 3
• Input 2 produces an output of 5

The function \( g(t) \) works the same way. Inputting values into the table allows us to find the exact outputs, assisting in understanding how the function operates over a range of values.

Composite Function

Composite functions are a fascinating concept where one function is nested inside another. This involves taking the output of one function and using it as the input for another.Given two functions, such as \( g(t) \) and another function \( g \), the composite function \( (g \circ g)(t) \) is formed. It can be thought of in steps:

• First, find \( g(t) \), which gives you a new value.
• Then, use this new value as the next input to the function \( g \).

This two-step process sometimes requires looking up two values in two tables consecutively, which is exactly what we did here.Our example finds composite functions like this:

• Let \( t = 1 \). First, find \( g(1) = 5 \).
• Then evaluate \( g(g(1)) = g(5) = 2 \), giving us \( (g \circ g)(1) = 2 \).

Understanding how to navigate through these layers of functions is crucial to mastering the concept of function composition.