Problem 19 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Suppose that the functions \(f, g\), and \(h\) are de ned for all integers. At the top of the following page is a table of some of the values of these functions. $$\begin{array}{lccccccc} \hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 0 & 2 & 1 & 3 & 4 & -2 & 5 \\ g(x) & 2 & 3 & 4 & 1 & 3 & -1 & 0 \\ h(x) & 3 & 4 & -3 & 2 & 8 & 1 & 2 \\ \hline \end{array}$$ Evaluate the following expressions. If not enough information is available for you to do so, indicate that. (a) \(f(-1) \cdot g(-1)\) (b) \(f(g(-1))\) (c) \(g(f(-1))\) (d) \(h(g(f(2)))\) (e) \(\frac{f(0)+2}{g(0)}\) (f) \(5 h(3)+f(f(1))\) (g) \(f(f(f(0)))\)

Step-by-Step Solution

Answer

From the respective steps, the results are: (a) 6, (b) -2, (c) 3, (d) -3, (e) 0.75, (f) 3 and (g) -2.

Step 1: Understanding the Table

In this table \( x \) is the input of the function, and the corresponding value in the row of the function is the output. For example, \( f(-1) = 2 \), \( g(-1) = 3 \), and \( h(-1) = 4 \).

Step 2: Evaluate \(f(-1) \cdot g(-1)\)

To find \(f(-1) \cdot g(-1)\), first find the values of \(f(-1)\) and \(g(-1)\) from the table which are 2 and 3 respectively. Multiply these together to get 6.

Step 3: Evaluate \(f(g(-1))\)

To find \(f(g(-1))\), first find the value of \(g(-1)\) from the table which is 3. Now find the value of \(f(3)\) from the same table, which is -2. Therefore, \(f(g(-1)) = -2\).

Step 4: Evaluate \(g(f(-1))\)

To evaluate \(g(f(-1))\), first find \(f(-1)\) from the table which is 2. Now find \(g(2)\) from the table, which is 3. Therefore, \(g(f(-1)) = 3\).

Step 5: Evaluate \(h(g(f(2)))\)

To find \(h(g(f(2)))\), first find \(f(2)\) from the table which is 4. Now find \(g(4)\) from the table which is 0. Finally, look for \(h(0)\) which is -3. Therefore, \(h(g(f(2))) = -3\).

Step 6: Evaluate \(\frac{f(0)+2}{g(0)}\)

First find \(f(0)\) from the table which is 1. Then, add 2 to this value which is 3. Now find \(g(0)\) from the table which is 4. Divide the two results to get \(\frac{3}{4}\). Therefore, \(\frac{f(0)+2}{g(0)} = \frac{3}{4}\).

Step 7: Evaluate \(5 h(3)+f(f(1))\)

To evaluate this, first find \(h(3)\) from the table which is 1. Multiply this by 5 to get 5. Now find \(f(1)\) from the table which is 3. Then find \(f(3)\) from the table which is -2. Therefore, \(5 h(3)+f(f(1)) = 5 + -2 = 3 \).

Step 8: Evaluate \(f(f(f(0)))\)

We first find \(f(0)\) from the table which is 1. Next, we find \(f(1)\) from the table which is 3. Finally, we find \(f(3)\) from the table which is -2. Therefore, \(f(f(f(0))) = -2\).

Key Concepts

Composite Functions

Composite functions are functions that result from combining two or more functions. They offer a way to build new functions by using the outputs of one function as the inputs to another.
In this context, when combining the functions provided in the table, you might see expressions like \(f(g(-1))\) or \(h(g(f(2)))\).
This means that you first find the output of the innermost function and use that as the input for the next one in line. For example, in \(h(g(f(2)))\):

• Start with \(f(2)\), find its value, which is 4.
• Use this result to find \(g(4)\), which gives 0.
• Finally, look up \(h(0)\), resulting in -3.

Composite functions often allow us to express complex problems more concisely and can help model real-world situations more effectively.

Function Composition

Function composition is the process of applying one function to the results of another. It is denoted as \( (f \circ g)(x) = f(g(x)) \), meaning you apply \(g\) first and then \(f\).
It's like a chain reaction where the output of one function feeds directly into another. This technique can simplify complex operations by breaking them down into simpler steps.
In the context of the provided solution, function composition is used to evaluate expressions like \(f(g(-1))\).
The calculation involves:

• First evaluating \(g(-1)\) to get 3.
• Then, using this to evaluate \(f(3)\), which yields -2.

This method is powerful in mathematical analysis, allowing you to manipulate and transform functions systematically, ultimately leading to more nuanced solutions.

Integer Functions

An integer function is a type of function whose inputs and outputs are restricted to integer values. This is evident in the table where each function, whether \(f, g,\) or \(h\), is defined for integer inputs like -2, -1, 0, etc.
These functions are particularly important in number theory and discrete mathematics. They provide clear and straightforward relationships without involving fractions or decimals.
In the problem context, integer functions were used consistently. For instance, finding \(f(f(f(0)))\) involved tracking integer outputs step-by-step:

• First, \(f(0) = 1\).
• Next, \(f(1) = 3\).
• Finally, \(f(3) = -2\).

Working with integer functions helps simplify mathematical calculations, making it easier to model and solve problems with whole numbers only.