Chapter 4: Problem 42
Find the inverse of $$y=4 x+2(5-x)$$
Short Answer
Expert verified
\(f^{-1}(x) = \frac{x - 10}{2}\)
Step by step solution
01
Simplify the given function
First, simplify the expression on the right side by distributing and combining like terms. The given function is: \(y = 4x + 2(5 - x)\) Distribute the 2 into the parentheses: \(y = 4x + 10 - 2x\) Combine like terms: \(y = (4x - 2x) + 10\) \(y = 2x + 10\)
02
Swap variables to find the inverse function
To find the inverse, swap the \(x\) and \(y\) in the equation. After swapping variables, the equation becomes: \(x = 2y + 10\)
03
Solve for the new \(y\)
Solve the equation from Step 2 for the new \(y\) to get the inverse function. Subtract 10 from both sides of the equation: \(x - 10 = 2y\) Divide both sides by 2: \(\frac{x - 10}{2} = y\) The inverse function is: \(y = \frac{x - 10}{2}\) or \(f^{-1}(x) = \frac{x - 10}{2}\)
04
Write the inverse function notation
Finally, express the solution with inverse function notation. The inverse of \(y = 2x + 10\) is written as: \(f^{-1}(x) = \frac{x - 10}{2}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving Equations
Solving equations is one of the foundational skills in algebra that allows us to find the value of unknown variables. The process involves a sequence of operations to isolate the variable on one side of the equation, often through the use of addition, subtraction, multiplication, or division.
Let's apply this to our problem with inverse functions. To solve for the new variable in the equation
Let's apply this to our problem with inverse functions. To solve for the new variable in the equation
x = 2y + 10
, we manipulated it to isolate y
. This involved subtraction followed by division — subtracting 10 from both sides to remove the constant term adjacent to 2y
, and then dividing both sides by 2 to find the value of y
. The result, y = (x - 10)/2
, is the inverse function which tells us what we need to input into the original function to get a desired output. Function Simplification
Function simplification involves reducing a function to its most basic form to make it easier to work with. This can involve expanding expressions, combining like terms, and reducing fractions when possible.
Using the initial function
Using the initial function
y = 4x + 2(5 - x)
, we first distributed the 2 across the parentheses and then combined the like terms of x
to simplify the function to y = 2x + 10
. Simplification is essential, as it can reveal the underlying relationship in a more understandable manner, making the next steps of finding inverses or solving equations less complicated. Inverse Function Notation
Inverse function notation is used to denote the function that reverses the effect of the original function. If we have a function
For instance, in our problem, we translated
f(x)
, the inverse function is commonly written as f^{-1}(x)
. This notation expresses the idea that if we apply f^{-1}
to a result of f(x)
, we would get back to our starting value of x
.For instance, in our problem, we translated
y = 2x + 10
into inverse function notation as f^{-1}(x) = (x - 10)/2
. Understanding this notation is crucial as it allows us to communicate and work with inverse functions effectively. Algebraic Manipulation
Algebraic manipulation refers to the process of rearranging, simplifying, or rewriting algebraic expressions in a more useful form. This skill is invaluable when working with equations and functions, as you often need to isolate variables, factor expressions, or expand parentheses.
In the example we are discussing, we began with distributing the 2 inside the brackets of
In the example we are discussing, we began with distributing the 2 inside the brackets of
y = 4x + 2(5 - x)
and then combined the resulting terms of x
to simplify the expression. Later, during the process of finding the inverse, we reversed the roles of x
and y
, and algebraically manipulated the expression to solve for the new y
. Mastering algebraic manipulation can make complex problem-solving far more approachable and allow students to tackle a wide range of algebraic challenges.