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Magazine Chapter 1: Problem 6
If \(f(x)=x-1\) and \(g(x)=1 /(x+1),\) find the following. $$ \begin{array}{ll}{\text { a. }} & {f(g(1 / 2))} & {\text { b. } g(f(1 / 2))} \\\ {\text { c. }} & {f(g(x))} & {\text { d. } g(f(x))} \\ {\text { e. }} & {f(f(2))} & {\text { f. } g(g(2))} \\ {\text { g. }} & {f(f(x))} & {\text { h. } g(g(x))}\end{array} $$
Short Answer
Expert verified
a) -1/3, b) 2, c) -x/(x+1), d) 1/x, e) 0, f) 3/4, g) x-2, h) (x+1)/(x+2).
Step by step solution
01
Evaluate g(1/2)
First, we need to find the value of \( g(1/2) \). \[ g(x) = \frac{1}{x+1} \] so substituting 1/2 for \( x \), we have \[ g(1/2) = \frac{1}{1/2 + 1} = \frac{1}{3/2} = \frac{2}{3}. \]
02
Evaluate f(g(1/2))
Next, substitute \( g(1/2) = \frac{2}{3} \) into \( f(x) = x - 1 \). So, \[ f\left(\frac{2}{3}\right) = \frac{2}{3} - 1 = \frac{2}{3} - \frac{3}{3} = -\frac{1}{3}. \] This is the solution for part (a).
03
Evaluate f(1/2)
Compute \( f(1/2) \) by substituting it into \( f(x) = x - 1 \): \[ f(1/2) = \frac{1}{2} - 1 = \frac{1}{2} - \frac{2}{2} = -\frac{1}{2}. \]
04
Evaluate g(f(1/2))
Now, use the result of Step 3 to find \( g(f(1/2)) \). Substitute \( -\frac{1}{2} \) into \( g(x) = \frac{1}{x+1} \): \[ g\left(-\frac{1}{2}\right) = \frac{1}{-\frac{1}{2} + 1} = \frac{1}{\frac{1}{2}} = 2. \] This is the solution for part (b).
05
Find the expression for f(g(x))
Substitute \( g(x) = \frac{1}{x+1} \) into \( f(x) = x - 1 \): \[ f(g(x)) = \frac{1}{x+1} - 1 = \frac{1}{x+1} - \frac{x+1}{x+1} = \frac{1-(x+1)}{x+1} = \frac{-x}{x+1}. \] This is the solution for part (c).
06
Find the expression for g(f(x))
Substitute \( f(x) = x - 1 \) into \( g(x) = \frac{1}{x+1} \): \[ g(f(x)) = \frac{1}{(x-1)+1} = \frac{1}{x}. \] This is the solution for part (d).
07
Evaluate f(2)
Compute \( f(2) \) by substituting it into \( f(x) = x - 1 \): \[ f(2) = 2 - 1 = 1. \]
08
Evaluate f(f(2))
Now use the result from step 7 to find \( f(f(2)) \). Substitute \( 1 \) into \( f(x) \): \[ f(1) = 1 - 1 = 0. \] This is the solution for part (e).
09
Evaluate g(2)
Compute \( g(2) \) by substituting it into \( g(x) = \frac{1}{x+1} \): \[ g(2) = \frac{1}{2+1} = \frac{1}{3}. \]
10
Evaluate g(g(2))
Now use the result from Step 9 to find \( g(g(2)) \). Substitute \( \frac{1}{3} \) into \( g(x) \): \[ g\left(\frac{1}{3}\right) = \frac{1}{\frac{1}{3} + 1} = \frac{1}{\frac{4}{3}} = \frac{3}{4}. \] This is the solution for part (f).
11
Find the expression for f(f(x))
Substitute \( f(x) = x - 1 \) back into itself: \[ f(f(x)) = f(x - 1) = (x - 1) - 1 = x - 2. \] This is the solution for part (g).
12
Find the expression for g(g(x))
Substitute \( g(x) = \frac{1}{x+1} \) back into itself: \[ g(g(x)) = \frac{1}{\left(\frac{1}{x+1}\right) + 1} = \frac{1}{\frac{1 + (x+1)}{x+1}} = \frac{x+1}{x+2}. \] This is the solution for part (h).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Function Evaluation
Function evaluation is all about finding the output of a function when we substitute a particular value for the variable in the function's equation. In this exercise, you have two functions, \( f(x) = x - 1 \) and \( g(x) = \frac{1}{x+1} \). To evaluate these functions, you substitute specific values for \( x \).
- Example: To find \( g(1/2) \), replace \( x \) with \( 1/2 \) in the expression for \( g(x) \), resulting in \( g(1/2) = \frac{2}{3} \).
- Similarly, for \( f(2) \), replace \( x \) with \( 2 \), which gives \( f(2) = 1 \).
Exploring Nested Functions
Nested functions occur when one function is placed inside another as its input. In this exercise, functions like \( f(g(x)) \) or \( g(f(x)) \) are examples of this.
- For \( f(g(x)) \), you first evaluate \( g(x) \), and then use that result within the function \( f \).
- This means after substituting into \( g(x) \), the output becomes the input for \( f(x) \). For example, if we know \( g(x) \) equals \( \frac{1}{x+1} \), that entire expression becomes what's inputted into \( f \).
Mastering Algebraic Manipulation
Algebraic manipulation involves restructuring or reorganizing algebraic expressions to solve equations or make them easier to work with. This exercise heavily relies on manipulating expressions to simplify nested functions.
- For instance, to simplify \( f(g(x)) = \frac{1}{x+1} - 1 \), you realize that \( 1 \) can be rewritten as \( \frac{x+1}{x+1} \) to have a common denominator, resulting in \( \frac{-x}{x+1} \).
- Similarly, converting \( g(g(x)) \) involves keeping track of fractions, where reorganizing terms enables you to express it as \( \frac{x+1}{x+2} \).
Grasping Rational Functions
A rational function is a function that can be written as the quotient of two polynomials. The function \( g(x) = \frac{1}{x+1} \) is a typical example of this.
- Rational functions can sometimes create undefined values, for instance, where the denominator becomes zero. So, \( g(x) \) is undefined at \( x = -1 \).
- Understanding such points is crucial, as they're often excluded from the function's domain.
