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Show that the given function is one-to-one and find its inverse. Check your answers algebraically and graphically. Verify that the range of \(f\) is the domain of \(f^{-1}\) and vice-versa. $$f(x)=3-\sqrt[3]{x-2}$$

Short Answer

Expert verified
The function is one-to-one, with inverse \( f^{-1}(x) = (3-x)^3 + 2 \).

Step by step solution

01

Determine Injectivity

To determine if the function \( f(x) = 3 - \sqrt[3]{x-2} \) is one-to-one, we need to check that different inputs produce different outputs. Assume \( f(a) = f(b) \). Then we have:\[ 3 - \sqrt[3]{a-2} = 3 - \sqrt[3]{b-2} \] Simplifying gives:\[ \sqrt[3]{a-2} = \sqrt[3]{b-2} \] Since the cube root is a one-to-one function, it follows that:\[ a-2 = b-2 \] Thus, \( a = b \), proving that \( f(x) \) is injective.
02

Find the Inverse Function

To find \( f^{-1}(x) \), set \( y = f(x) = 3 - \sqrt[3]{x-2} \) and solve for \( x \) in terms of \( y \):\[ y = 3 - \sqrt[3]{x-2} \] \[ \sqrt[3]{x-2} = 3 - y \]By cubing both sides, we get:\[ x - 2 = (3 - y)^3 \] Solving for \( x \), we find:\[ x = (3 - y)^3 + 2 \] Thus, the inverse function is \( f^{-1}(x) = (3-x)^3 + 2 \).
03

Verify Algebraically

To verify, compute \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \):1. Compute \( f(f^{-1}(x)) \): \[ f((3-x)^3 + 2) = 3 - \sqrt[3]{(3-x)^3 + 2 - 2} = 3 - \sqrt[3]{(3-x)^3} = 3 - (3-x) = x \]2. Compute \( f^{-1}(f(x)) \): \[ f^{-1}(3 - \sqrt[3]{x-2}) = ((3 - (3 - \sqrt[3]{x-2}))^3 + 2 = (\sqrt[3]{x-2})^3 + 2 = x - 2 + 2 = x \]Both results lead to \( x \), confirming the inverse correctly.
04

Verify Graphically

Graph \( f(x) = 3 - \sqrt[3]{x-2} \) and \( f^{-1}(x) = (3-x)^3 + 2 \). The graphs should be symmetric across the line \( y = x \). This symmetry visually confirms that one is the inverse of the other.
05

Verify Domains and Ranges

The domain of \( f(x) \) is all real numbers \( x \). The range is all real numbers \( y \), since the cube root function covers all real numbers. Thus, the domain of \( f^{-1}(x) \) is also all real numbers in accordance with the range of \( f(x) \), and vice versa. This confirms our functions are correctly determined.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-to-One Functions
A one-to-one function is a special kind of function where each input is paired with a unique output. This means that no two different inputs in the domain can produce the same output.
For a function to be one-to-one, it must be injective.
  • If we assume that two outputs, say \( f(a) \) and \( f(b) \), are equal, it should logically follow that \( a = b \) for the function to be one-to-one.
  • The proof uses this property to show uniqueness or distinctness of outputs relative to inputs.
In this particular exercise, we examine the function \( f(x) = 3 - \sqrt[3]{x-2} \). By setting \( f(a) = f(b) \) and simplifying, we see that \( a = b \). Thus, the function is one-to-one because different \( x \) values cannot produce the same output. This property is crucial when finding inverses because only one-to-one functions have well-defined inverses.
Cubic Functions
Cubic functions are polynomial functions of degree three and follow the form \( f(x) = ax^3 + bx^2 + cx + d \). In cubic functions, there is typically one turning point and up to three roots or solutions.
  • The function in our exercise, \( f(x) = 3 - \sqrt[3]{x-2} \), incorporates a cube root operation, connecting it with cubic behavior.
  • The inverse nature of cube roots is key in the process of finding the inverse function.
When you encounter cube roots like in our exercise, remember that they span all real numbers and are symmetric about the origin due to their inherent nature. This symmetry aids in understanding their graphs and inverses.
Domain and Range
The domain and range are fundamental aspects of any function, defining where it operates and what outputs are possible.
  • The domain represents all possible inputs (\( x\) values), while the range describes all possible outputs (\( y\) values).
  • For \( f(x) = 3 - \sqrt[3]{x-2} \), the domain includes all real numbers because the cube root function, \( \sqrt[3]{x}\), can handle all real numbers.
  • Similarly, the range is also all real numbers because the function subtracts a cube root from a constant, allowing it to reach any real number.
In this exercise, the domain of \( f\) becomes the range of \( f^{-1}\), and vice versa, which exemplifies the natural behavior of inverse functions.
Function Verification
Verification of functions, particularly inverse functions, ensures that they are accurately calculated and behave as expected.
  • For inverse verification, algebraic manipulation can show that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) satisfy the properties of inverse functions.
  • Graphical verification gives a visual check; if the function and its inverse reflect over the line \( y = x \), they are accurately inverse functions of each other.
In our exercise, both algebraic and graphical methods confirm the accuracy of \( f^{-1}(x) = (3-x)^3 + 2 \) as the inverse of \( f(x) = 3 - \sqrt[3]{x-2} \). This dual verification reinforces confidence in our findings.

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