Problem 13 - Algebra 2

Question

Find the inverse of each function. Is the inverse a function? $$ y=(1-2 x)^{2}+5 $$

Step-by-Step Solution

Answer

The inverse of the given function are \(y=\frac{1-\sqrt{x-5}}{2}\) and \(y=\frac{1+\sqrt{x-5}}{2}\). However, since the inverse is not a function, there is a limitation for the given exercise.

Step 1: Interchange x and y

Starting from the equation \(y=(1-2x)^2 + 5\). The roles of \(x\) and \(y\) will be interchanged to get \(x=(1-2y)^2 + 5\).

Step 2: Solve for y

Subtract 5 from both sides to obtain \((1-2y)^2 = x - 5\). Take the square root of both sides to get \(1-2y = \sqrt{x-5}\) or \(1-2y= -\sqrt{x-5}\). Finally, solve for \(y\) to give \(y=\frac{1-\sqrt{x-5}}{2}\) and \(y=\frac{1+\sqrt{x-5}}{2}\).

Step 3: Test for function

The inverse of the original function gives two outputs for one input. Hence, the inverse of the given function is not a function.

Key Concepts

Function Properties

Functions are mathematical relationships where each input corresponds to exactly one output. This rule ensures that for every domain value, there is a unique range value, which is often what differentiates functions from general relations. In the original exercise, we start with a function and are asked to determine its inverse. A crucial property of functions is whether their inverses also satisfy the condition of one output per input.
To check if the inverse of a function is indeed a function, we ensure the inverse still maintains this uniqueness. In the case of the given function \( y = (1-2x)^2 + 5 \), we find in the final step that the inverse does not satisfy this essential property, having two possible outputs for a single input.

Square Root

The square root is a mathematical operation that finds the value that, when multiplied by itself, gives the original number. In the process of finding inverses, taking the square root is a common step, especially when dealing with quadratic functions.
From the equation, \( (1-2y)^2 = x - 5 \), applying the square root requires careful attention to both the positive and negative roots. This is because any positive number has two square roots: the positive and the negative square root. Thus, the expression becomes \( 1-2y = \sqrt{x-5} \) or \( 1-2y = -\sqrt{x-5} \). Recognizing both possibilities is essential since they impact whether the inverse can consistently provide unique outputs, which influences its status as a function.

Solving Equations

Solving for a variable involves manipulating an equation to express one variable solely in terms of others. This is a vital skill in determining inverses.
In this context, we modify the expression \( x = (1-2y)^2 + 5 \) to solve for \( y \). First, subtract 5 from both sides, resulting in \( (1-2y)^2 = x - 5 \). Next, the square root operation is applied, considering both positive and negative roots, giving two equations: \( 1-2y = \sqrt{x-5} \) and \( 1-2y = -\sqrt{x-5} \). By solving these, we get \( y = \frac{1-\sqrt{x-5}}{2} \) and \( y = \frac{1+\sqrt{x-5}}{2} \), respectively. Each solution reveals different inverse possibilities, which we use to test if the inverse is a function.

Domain and Range

In a function, the domain is the set of all possible input values, while the range is the set of all possible outputs. When finding an inverse, these roles switch: the domain of the inverse is the range of the original function, and vice versa.
This becomes complex particularly when dealing with square roots due to natural constraints, like not taking the square root of a negative number. Here, for \( x = (1-2y)^2 + 5 \), the domain of the inverse involves values where \( x-5 \geq 0 \) as you can only take the square root of non-negative numbers. Consequently, \( x \geq 5 \) for the domain. Similarly, the concept of range when finding inverses deals with whether negative roots affect output consistency, influencing if the inverse is a function.