Chapter 7: Problem 22
Gives a formula for a function \(y=f(x)\) and shows the graphs of \(f\) and \(f^{-1} .\) Find a formula for \(f^{-1}\) in each case. $$f(x)=x^{2}-2 x+1, \quad x \geq 1$$
Short Answer
Expert verified
The inverse function is \( f^{-1}(x) = \sqrt{x} + 1 \) for \( x \geq 0 \).
Step by step solution
01
Understand the Problem
We are given a function \( f(x) = x^2 - 2x + 1 \) and we need to find its inverse, \( f^{-1}(x) \). Since \( x \geq 1 \), this restricts the domain of the function allowing us to find an inverse.
02
Simplify the Function
Start by looking at the function: \( f(x) = x^2 - 2x + 1 \). Notice that this can be rewritten as \( (x-1)^2 \). Therefore, \( f(x) = (x-1)^2 \).
03
Set Up the Inverse Function
To find \( f^{-1}(x) \), we need to solve the equation \( y = (x-1)^2 \) for \( x \), giving us the inverse in terms of \( y \).
04
Solve the Equation for Inverse
Start by expressing the variable \( x \) in terms of \( y \). We have \( y = (x-1)^2 \). By taking the square root of both sides, we get \( \sqrt{y} = x-1 \). Solving for \( x \), we obtain \( x = \sqrt{y} + 1 \).
05
Write the Final Expression for Inverse Function
Thus, the inverse function is \( f^{-1}(x) = \sqrt{x} + 1 \), valid for \( x \geq 0 \) due to the square root definition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Finding Inverse Functions
To find the inverse of a function means determining another function that `undoes` the effect of the original. For a function \( f(x) \), its inverse \( f^{-1}(x) \) switches the roles of inputs (x) and outputs (y). This implies that if \( f(a) = b \), then \( f^{-1}(b) = a \). Finding the inverse involves a series of steps:
- First, express the function \( y = f(x) \) in terms of y.
- Next, solve for \( x \) in terms of \( y \). This involves algebraic manipulation and sometimes taking square roots or other operations.
- Finally, swap the variables \( x \) and \( y \) to get \( f^{-1}(x) \).
Domain and Range in Inverse Functions
The domain and range are crucial concepts when dealing with functions and their inverses. The **domain** of a function is the set of all possible input values (x-values), while the **range** is the set of possible output values (y-values). When finding the inverse of a function, the domain and range switch places.
For our given function \( f(x) = x^2 - 2x + 1 \), where \( x \geq 1 \), the domain is restricted to \( x \geq 1 \). The form \( f(x) = (x-1)^2 \) indicates that the range starts from 0 because \((x-1)^2\) cannot be negative. Thus, the range of \( f(x) \) is \( y \geq 0 \).
Inverting the function to \( f^{-1}(x) = \sqrt{x} + 1 \), the domain of the inverse becomes what the range of the original was: \( x \geq 0 \). Conversely, the range of the inverse is \( y \geq 1 \), as the smallest value \( \sqrt{x} + 1 \) can take is 1 (when \( x = 0 \)). Understanding these swapped roles assists in accurately identifying and working with inverse functions.
For our given function \( f(x) = x^2 - 2x + 1 \), where \( x \geq 1 \), the domain is restricted to \( x \geq 1 \). The form \( f(x) = (x-1)^2 \) indicates that the range starts from 0 because \((x-1)^2\) cannot be negative. Thus, the range of \( f(x) \) is \( y \geq 0 \).
Inverting the function to \( f^{-1}(x) = \sqrt{x} + 1 \), the domain of the inverse becomes what the range of the original was: \( x \geq 0 \). Conversely, the range of the inverse is \( y \geq 1 \), as the smallest value \( \sqrt{x} + 1 \) can take is 1 (when \( x = 0 \)). Understanding these swapped roles assists in accurately identifying and working with inverse functions.
Quadratic Functions
Quadratic functions are equations of the form \( ax^2 + bx + c \) and are characterized by their **parabolic** shapes. A key property is that they often include a vertex, the highest or lowest point on the graph, depending on the orientation. These functions can sometimes be rewritten in vertex form: \( a(x-h)^2 + k \), which reveals the vertex \( (h, k) \).
For the quadratic function \( f(x) = x^2 - 2x + 1 \), recognizing it in its transformed form \( (x-1)^2 \) indicates it's a perfect square trinomial. This makes finding the inverse easier as we directly express the transformation and find \( f^{-1} (x) = \sqrt{x} + 1 \).
Quadratic functions are **not automatically invertible** across their entire domain because they are not one-to-one (a one-to-one function has a unique x-value for each y-value). However, by restricting the domain, such as \( x \geq 1 \) in our example, these functions can have an inverse. These features highlight the versatility and distinct behavior of quadratic functions in algebra.
For the quadratic function \( f(x) = x^2 - 2x + 1 \), recognizing it in its transformed form \( (x-1)^2 \) indicates it's a perfect square trinomial. This makes finding the inverse easier as we directly express the transformation and find \( f^{-1} (x) = \sqrt{x} + 1 \).
Quadratic functions are **not automatically invertible** across their entire domain because they are not one-to-one (a one-to-one function has a unique x-value for each y-value). However, by restricting the domain, such as \( x \geq 1 \) in our example, these functions can have an inverse. These features highlight the versatility and distinct behavior of quadratic functions in algebra.