Given the function:
[tex]f(x)=x^2-2 x[/tex]
How can you restrict the domain so that [tex]f(x)[/tex] has an inverse? What is the equation of the inverse function?
A. [tex]x \geq 0 ; f^{-1}(x)=1+\sqrt{x+1}[/tex]
B. [tex]x \geq 1 ; f^{-1}(x)=1+\sqrt{x+1}[/tex]
C. [tex]x \geq 0 ; f^{-1}(x)=1-\sqrt{x+1}[/tex]
D. [tex]x \geq 1 ; f^{-1}(x)=1-\sqrt{x+1}[/tex]
Answer (1)
Find the vertex of the parabola f ( x ) = x 2 − 2 x , which is at x = 1 .
Restrict the domain to xg e 1 to make the function one-to-one.
Find the inverse function by swapping x and y and solving for y , resulting in y = 1 ± x + 1 .
Choose the positive square root since the range must be y g e 1 , so the inverse function is x ≥ 1 ; f − 1 ( x ) = 1 + x + 1 .
Explanation
Problem Analysis We are given the function f ( x ) = x 2 − 2 x and asked to find a suitable domain restriction so that f ( x ) has an inverse. We also need to find the equation of the inverse function.
Finding the Vertex First, let's find the vertex of the parabola f ( x ) = x 2 − 2 x . The x-coordinate of the vertex is given by x v = − b / ( 2 a ) , where a = 1 and b = − 2 . Thus, x v = − ( − 2 ) / ( 2 ∗ 1 ) = 1 . The y-coordinate of the vertex is f ( 1 ) = 1 2 − 2 ( 1 ) = − 1 . So the vertex is at ( 1 , − 1 ) .
Restricting the Domain Since the parabola opens upwards, we can restrict the domain to x ≥ 1 or x ≤ 1 to make the function one-to-one and thus invertible.
Finding the Inverse (Case 1) Case 1: Restrict the domain to x ≥ 1 . To find the inverse, let y = x 2 − 2 x . Swap x and y to get x = y 2 − 2 y .
Solving for y (Case 1) Solve for y : y 2 − 2 y − x = 0 . Complete the square: y 2 − 2 y + 1 = x + 1 , so ( y − 1 ) 2 = x + 1 . Taking the square root, y − 1 = ± x + 1 , so y = 1 ± x + 1 .
Determining the Correct Root (Case 1) Since we restricted the domain to x ≥ 1 , the range of the inverse must also be y ≥ 1 . Therefore, we choose the positive square root: f − 1 ( x ) = 1 + x + 1 . The domain of the inverse is x ≥ − 1 .
Finding the Inverse (Case 2) Case 2: Restrict the domain to x ≤ 1 . Following the same steps as in Case 1, we have y = 1 ± x + 1 .
Determining the Correct Root (Case 2) Since we restricted the domain to x ≤ 1 , the range of the inverse must also be y ≤ 1 . Therefore, we choose the negative square root: f − 1 ( x ) = 1 − x + 1 . The domain of the inverse is x ≥ − 1 .
Final Answer Comparing the obtained inverse functions with the given options, we see that the option x ≥ 1 ; f − 1 ( x ) = 1 + x + 1 is correct.
Examples
Understanding inverse functions is crucial in many scientific and engineering applications. For example, if you know the distance an object has fallen under gravity as a function of time, the inverse function will tell you how long it took to fall a certain distance. This concept is also used in cryptography, where encoding and decoding messages rely on inverse functions.
