Function inverses example 3 | Functions and their graphs | Algebra II | Khan Academy
TLDRThe video script discusses the process of finding the inverse of a function defined as f(x) = (x - 1)^2 - 2, with a domain constraint of x β€ 1, resulting in a half-parabola. The instructor guides through setting y = f(x) and solving for x in terms of y, with attention to the domain and range. Emphasis is placed on taking the negative square root due to the domain constraint, leading to the inverse function f^(-1)(y) = -β(y + 2 + 1) for y β₯ -2. The script concludes with a graphical representation of the inverse function, illustrating its reflection across the line y = x.
Takeaways
- π The function given is f(x) = (x - 1)^2 - 2, which represents the left half of a parabola.
- π« The domain of x is constrained to x β€ 1, meaning we are only considering the left side of the parabola.
- π To find the inverse function, we start by setting y = f(x) and then solve for x in terms of y.
- π The process involves switching the roles of x and y, and considering the original range as the new domain.
- β By adding 2 to both sides of the equation, we get y + 2 = (x - 1)^2.
- π’ The y constraint is y β₯ -2, which becomes the domain for the inverse function.
- π‘ A key insight is recognizing that we need to take the negative square root because x - 1 is negative within the given domain.
- π The negative square root is crucial to correctly solve for x and avoid confusion with the positive square root.
- π The inverse function is found to be f^(-1)(y) = -β(y + 2 + 1) for y β₯ -2.
- π The graph of the inverse function is a reflection of the original function across the line y = x.
- π€ This problem is considered challenging, especially for a precalculus class, due to the need to understand the concept of taking the negative square root.
Q & A
What is the given function f(x)?
-The given function f(x) is f(x) = (x - 1)^2 - 2, with the domain constrained to x β€ 1.
Why is finding the inverse of the given function difficult?
-Finding the inverse is difficult because the domain is constrained to x β€ 1, which means we are dealing with only half of a parabola, not the full 'U' shape, complicating the process of finding the inverse function.
What is the initial step to find the inverse function?
-The initial step is to set y equal to f(x), or y = (x - 1)^2 - 2, and then solve for x in terms of y to find the inverse.
What is the domain of the original function f(x)?
-The domain of the original function f(x) is x β€ 1.
What is the range of the original function that will become the domain of the inverse function?
-The range of the original function, which becomes the domain of the inverse function, is y β₯ -2.
What is the first manipulation done to the equation to solve for x?
-The first manipulation is to add 2 to both sides of the equation to isolate the squared term, resulting in y + 2 = (x - 1)^2.
Why is it necessary to consider the negative square root when solving for x?
-It is necessary to consider the negative square root because the domain constraint (x β€ 1) implies that (x - 1) is negative, and we need to preserve the negative sign when taking the square root to solve for x.
What is the final expression for the inverse function f^(-1)(y)?
-The final expression for the inverse function is f^(-1)(y) = -β(y + 2 + 1) for y β₯ -2.
How does the graph of the inverse function compare to the original function's graph?
-The graph of the inverse function is a reflection of the original function's graph along the line y = x, due to the swapping of x and y.
What is the significance of taking the negative square root in the context of the domain constraint?
-Taking the negative square root is significant because it ensures that the solution for x remains within the constrained domain of x β€ 1, as (x - 1) is negative in this domain.
Outlines
π Finding the Inverse of a Constrained Parabola
The script begins by introducing a mathematical function, f(x) = (x - 1)^2 - 2, with a constrained domain of x β€ 1, creating a half-parabola. The challenge is to find the inverse of this function, which is complicated by the domain restriction. The process involves setting y = f(x) and then solving for x in terms of y, while also considering the range of y, which is β₯ -2. The speaker guides through the algebraic steps, emphasizing the importance of domain and range, and the need to take the negative square root due to the domain constraint, leading to the inverse function f^(-1)(x) = -β(x + 2 + 1) for x β₯ -2.
π Graphing the Inverse Function and Its Properties
The second paragraph continues the discussion on finding the inverse function, now focusing on graphing it. The speaker simplifies the inverse function to f^(-1)(x) = -β(x + 3) for x β₯ -2 and renames y as x to clarify the function in terms of x. The paragraph includes a step-by-step analysis of specific points on the graph, such as (-2, 1), (-1, 0), and (2, -1), to illustrate the shape of the inverse function. The speaker concludes by noting that the graph of the inverse function is a reflection of the original function across the line y = x, highlighting the difficulty of this precalculus problem and the significance of correctly choosing the negative square root due to the domain constraint.
Mindmap
Keywords
π‘Function
π‘Domain
π‘Parabola
π‘Inverse Function
π‘Square Root
π‘Negative Square Root
π‘Range
π‘Graph
π‘Constrain
π‘Solve for x
π‘Reflection
Highlights
The function f(x) is defined as x minus 1 squared minus 2 with a constrained domain of x β€ 1, representing the left half of a parabola.
The challenge of finding the inverse of the function due to the constrained domain is introduced.
The process begins by setting y equal to f(x), or y = x - 1 squared minus 2.
The goal is to express x in terms of y to find the inverse function.
The graph's range is identified as y β₯ -2, which will become the domain after swapping x and y.
The equation y + 2 = (x - 1) squared is derived by adding 2 to both sides.
The importance of considering the negative square root due to the domain constraint is emphasized.
The negative square root is taken to solve for x, resulting in x - 1.
The inverse function is expressed as f^-1(y) = -β(y + 2 + 1) for y β₯ -2.
Renaming y as x in the inverse function gives f^-1(x) = -β(x + 2 + 1) for x β₯ -2.
The graph of the inverse function is described as a reflection of f(x) along the line y = x.
The inverse function's graph is illustrated with specific points such as (-2, 1), (-1, 0), and (2, -1).
The complexity of the inverse problem in precalculus due to the need to take the negative square root is highlighted.
The importance of domain constraints in determining the correct square root to take is discussed.
The process of solving for the inverse function is summarized, emphasizing the steps taken.
The final expression for the inverse function and its domain are clearly stated.
The practical application of understanding domain and range when finding inverse functions is illustrated.
The transcript concludes with a reflection on the difficulty and educational value of the inverse problem presented.
Transcripts
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