Problem 33
Question
Each of Exercises \(25-34\) gives a formula for a function \(y=f(x)\) . In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1}\) . As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x.\) $$f(x)=x^{2}-2 x, \quad x \leq 1\quad \text (Hint: Complete \ the \ square.)$$
Step-by-Step Solution
Verified Answer
The inverse function is \( f^{-1}(x) = \sqrt{x+1} + 1 \). Domain: \( x \geq -1 \). Range: \( y \leq 1 \).
1Step 1: Rewrite the function using Complete the Square
The given function is \( f(x) = x^2 - 2x \). To complete the square, rewrite it as \( (x^2 - 2x + 1) - 1 = (x-1)^2 - 1 \). So, the function becomes \( f(x) = (x-1)^2 - 1 \).
2Step 2: Identify the inverse function formula
To find the inverse function, first set \( y = f(x) = (x-1)^2 - 1 \). Switch \( x \) and \( y \) to solve for \( y \), giving \( x = (y-1)^2 - 1 \).
3Step 3: Solve for the inverse function
From \( x = (y-1)^2 - 1 \), add 1: \( x + 1 = (y-1)^2 \). Then, take the square root: \( \sqrt{x + 1} = y - 1 \). Therefore, \( y = \sqrt{x + 1} + 1 \). The inverse function is \( f^{-1}(x) = \sqrt{x + 1} + 1 \).
4Step 4: Determine the domain of the inverse function
The inverse function \( f^{-1}(x) = \sqrt{x + 1} + 1 \) is defined for \( x + 1 \geq 0 \), which means \( x \geq -1 \). Thus, the domain of \( f^{-1} \) is \([-1, \infty)\).
5Step 5: Determine the range of the inverse function
Since the range of the original function \( f(x) \) is \((-\infty, 1] \), and \( f(x) \) is decreasing since \( x \leq 1 \), the range of \( f^{-1}(x) \) corresponds to the domain of \( f(x) \), which is \((-fty, 1] \).
6Step 6: Verify the inverse functions
Verify by substitution: 1. Substitute \( f^{-1}(x) \) into \( f(x) \): \[ f(\sqrt{x+1} + 1) = ((\sqrt{x+1} + 1)-1)^2 - 1 = x \]2. Substitute \( f(x) \) into \( f^{-1}(x) \): \[ f^{-1}((x-1)^2 - 1) = \sqrt{((x-1)^2 - 1) + 1} + 1 = x \]This confirms \( f(f^{-1}(x)) = f^{-1}(f(x)) = x \).
Key Concepts
Domain and RangeComplete the SquareFunction Transformation
Domain and Range
The domain of a function is the set of all possible inputs or "x" values, while the range is the set of all possible outputs or "y" values. When dealing with inverse functions, these roles swap.
For the inverse function, which we found to be \( f^{-1}(x) = \sqrt{x+1} + 1 \), the domain is determined by the requirement that the expression under the square root needs to be non-negative. Thus, we have \( x + 1 \geq 0 \), giving us the domain \([-1, \infty)\).
The range of the inverse function \( f^{-1} \) is based on the domain of the original function \( f(x) \). Since the original function is \(f(x) = (x-1)^2 - 1\) and is defined for \( x \leq 1 \), its range finds its minimum at \( y = -1 \) (when \(x = 1\)) and goes up to infinity indirectly. Hence, the range of the inverse function is \((-\infty, 1]\).
Understanding how to map these domains and ranges requires familiarity with the function's behavior. Once mastered, you can easily determine these properties for similar inverse functions.
For the inverse function, which we found to be \( f^{-1}(x) = \sqrt{x+1} + 1 \), the domain is determined by the requirement that the expression under the square root needs to be non-negative. Thus, we have \( x + 1 \geq 0 \), giving us the domain \([-1, \infty)\).
The range of the inverse function \( f^{-1} \) is based on the domain of the original function \( f(x) \). Since the original function is \(f(x) = (x-1)^2 - 1\) and is defined for \( x \leq 1 \), its range finds its minimum at \( y = -1 \) (when \(x = 1\)) and goes up to infinity indirectly. Hence, the range of the inverse function is \((-\infty, 1]\).
Understanding how to map these domains and ranges requires familiarity with the function's behavior. Once mastered, you can easily determine these properties for similar inverse functions.
Complete the Square
Completing the square is a useful technique to simplify expressions and solve quadratics. Let's break it down further.
Given a quadratic expression like \( x^2 - 2x \), completing the square means rewriting it in the form \((x-h)^2 + k\). This form helps simplify solving quadratic equations and finding their inverses.
To complete the square for \( f(x) = x^2 - 2x \), follow these steps:
Given a quadratic expression like \( x^2 - 2x \), completing the square means rewriting it in the form \((x-h)^2 + k\). This form helps simplify solving quadratic equations and finding their inverses.
To complete the square for \( f(x) = x^2 - 2x \), follow these steps:
- Add and subtract the square of half of the coefficient of \( x \): the coefficient is \(-2\), half of \(-2\) is \(-1\), and \((-1)^2 = 1\).
- Rewrite \( x^2 - 2x \) as \((x^2 - 2x + 1) - 1 = (x - 1)^2 - 1\).
Function Transformation
Function transformation involves shifting, reflecting, stretching, or compressing the graph of a function. Understanding these transformations helps in graphing and solving equations.
The function \( f(x) = (x-1)^2 - 1 \) demonstrates transformation through its completed square form.
The function \( f(x) = (x-1)^2 - 1 \) demonstrates transformation through its completed square form.
- \((x-1)^2\) indicates a horizontal shift of 1 unit to the right.
- The \(-1\) outside means the entire graph is shifted 1 unit down.
- There is no vertical stretch or compression as the coefficient of \(x^2\) is 1.
Other exercises in this chapter
Problem 33
Evaluate the integrals. \begin{equation}\int\left(e^{3 x}+5 e^{-x}\right) d x\end{equation}
View solution Problem 33
In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln (\sec (\ln \theta)) $$
View solution Problem 34
In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\tan ^{-1}(\ln x) $$
View solution Problem 34
In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\operatorname{csch}^{-1} 2^{\theta}$$
View solution