Problem 32
Question
Determine whether each pair of functions are inverse functions. $$ \begin{array}{l}{f(x)=6 x+2} \\ {g(x)=x-\frac{1}{3}}\end{array} $$
Step-by-Step Solution
Verified Answer
Functions are not inverse; neither composition yields x.
1Step 1: Define Inverse Functions
Two functions \( f \) and \( g \) are inverses of each other if and only if \( f(g(x)) = x \) and \( g(f(x)) = x \) for all \( x \) in the domains of \( f \) and \( g \). We need to check both compositions to determine if \( f(x) \) and \( g(x) \) are inverse functions.
2Step 2: Compute f(g(x))
Substitute \( g(x) = x - \frac{1}{3} \) into \( f(x) = 6x + 2 \):\[ f(g(x)) = f(x - \frac{1}{3}) = 6(x - \frac{1}{3}) + 2 = 6x - 2 + 2 = 6x. \]The result is \( 6x \), not \( x \), which already suggests they are not inverse functions.
3Step 3: Compute g(f(x))
Substitute \( f(x) = 6x + 2 \) into \( g(x) = x - \frac{1}{3} \):\[ g(f(x)) = g(6x + 2) = (6x + 2) - \frac{1}{3} = 6x + 2 - \frac{1}{3}. \]To simplify, convert \( 2 \) to \( \frac{6}{3} \):\[ 6x + \frac{6}{3} - \frac{1}{3} = 6x + \frac{5}{3}. \]The result is \( 6x + \frac{5}{3} \), not \( x \), confirming they are not inverses.
Key Concepts
Function CompositionAlgebraic FunctionsDomain of a Function
Function Composition
In mathematics, function composition involves combining two functions so that the output of one function becomes the input of the other. This is a powerful concept that lets us investigate the relationships between functions. If you have two functions, say \( f(x) \) and \( g(x) \), you can create a new function by applying \( g \) to \( x \) and then applying \( f \) to the result, denoted as \( f(g(x)) \). The same idea applies in the reverse order, creating \( g(f(x)) \).
Function composition is not always commutative, which means \( f(g(x)) \) might not equal \( g(f(x)) \). When dealing with inverse functions, we rely on function composition because two functions are inverses if both \( f(g(x)) = x \) and \( g(f(x)) = x \) for every \( x \) in their domains.
Function composition is not always commutative, which means \( f(g(x)) \) might not equal \( g(f(x)) \). When dealing with inverse functions, we rely on function composition because two functions are inverses if both \( f(g(x)) = x \) and \( g(f(x)) = x \) for every \( x \) in their domains.
- Start with \( f(g(x)) \): Plug the formula for \( g(x) \)--\( x - \frac{1}{3} \) into \( f(x) = 6x + 2 \). This simplifies to \( 6x \), which is not equal to \( x \).
- Next, check \( g(f(x)) \): Use \( f(x) = 6x + 2 \) in \( g(x) = x - \frac{1}{3} \). The result, \( 6x + \frac{5}{3} \), is also not equal to \( x \).
- Thus, we verify through composition that the functions provided are not inverses.
Algebraic Functions
Algebraic functions are equations involving algebraic expressions made up of variables, numbers, and arithmetic operations such as addition, subtraction, multiplication, and division. These functions are central to many areas in math where we study their properties, like how they can be composed or combined.
An algebraic function can appear straightforward or more complex. For instance, \( f(x) = 6x + 2 \) and \( g(x) = x - \frac{1}{3} \) are algebraic because they involve simple operations on \( x \). Evaluating whether two algebraic functions are inverses involves substituting one function into the other and ensuring that the output simplifies to just \( x \).
An algebraic function can appear straightforward or more complex. For instance, \( f(x) = 6x + 2 \) and \( g(x) = x - \frac{1}{3} \) are algebraic because they involve simple operations on \( x \). Evaluating whether two algebraic functions are inverses involves substituting one function into the other and ensuring that the output simplifies to just \( x \).
- The process involves algebraic manipulation: solving equations step-by-step as in function composition.
- Even simple algebraic functions require careful substitution and simplification to uncover their relationship.
- This highlights the importance of clarity and accuracy in algebraic operations when assessing functions.
Domain of a Function
Understanding the domain of a function is crucial since it tells us what values the function can accept as input. The domain is the set of all possible \( x \)-values that can be plugged into the function without causing any undefined operations, like division by zero or the square root of a negative number.
In our exercise, the domain isn't explicitly restricted due to the simplicity of the linear forms \( f(x) = 6x + 2 \) and \( g(x) = x - \frac{1}{3} \), both of which can accept all real numbers as inputs. This makes function composition easier since there are no restrictions on potential \( x \) values to check.
In our exercise, the domain isn't explicitly restricted due to the simplicity of the linear forms \( f(x) = 6x + 2 \) and \( g(x) = x - \frac{1}{3} \), both of which can accept all real numbers as inputs. This makes function composition easier since there are no restrictions on potential \( x \) values to check.
- The function \( f(x) \) is defined for all real \( x \), thus having an infinite domain.
- Likewise, function \( g(x) \) doesn't impose restrictions, allowing for a comprehensive evaluation of its range.
- When exploring inverses, having a wide domain facilitates checking every necessary instance where \( f(g(x)) = x \) or \( g(f(x)) = x \).
Other exercises in this chapter
Problem 32
Simplify. $$ \sqrt{36 x^{4} z^{4}} $$
View solution Problem 32
Describe the difference between the graphs of \(y=\sqrt{x}-4\) and \(y=\sqrt{x-4}\)
View solution Problem 32
Find \([g \circ h](x)\) and \([h \circ g](x)\) $$ \begin{array}{l}{g(x)=2 x} \\ {h(x)=x^{3}+x^{2}+x+1}\end{array} $$
View solution Problem 33
Solve each inequality. $$ -2+\sqrt{9-5 x} \geq 6 $$
View solution