Suppose that y = f(x) = x^2 - 4x + 4.
Then on any interval where the inverse function y = f-1(x) exists, thederivative of y = f-1(x) with respect to x is:
(Hint: x^2 - 4x + 4 can be factored and rewritten as "something" squared.)
Work thus far:
Okay, so I know that I can factor y to equal y = (x - 2)^2
Next, I can solve for the inverse function:
x = (y - 2)^2
sqrt(x) = y - 2
y = sqrt(x) + 2
Then, I can find the derivative:
y' = (1/2)x^(-1/2)
Now, my concern is that the question also hints to an implicit solution to this problem. Working out the problem implicitly, I get:
x = y^2 - 4x + 4
1 = 2y(dy/dx) - 4(dy/dx)
dy/dx = 1 / (2y - 4) where x and y satisfy the equation y = x^2 - 4x + 4.
Would this answer be correct, as well? Or should it be "where x and y satisfy the equation x = y^2 - 4y +4"?
The answer is 1/2x^-1/2
from y = f(x) = x^2 - 4x + 4
forming the inverse consists of interchanging the x and y variables, so the inverse is
x = y^2 - 4y + 4
(you had -4x instead of -4y, but apparently worked it as -4y )
so 1 = 2y dy/dx - 4dy/dx
dy/dx(2y-4) = 1
dy/dx of the inverse is 1/(2y-4)
but your restriction would be y ≠ 2
^thats wrong. I got it wrong. sorry.
To find the derivative of the inverse function, we can begin by finding the inverse function of y = f(x) = x^2 - 4x + 4.
First, we recognize that the expression x^2 - 4x + 4 can be factored as (x - 2)^2. Therefore, we can rewrite f(x) as y = (x - 2)^2.
To find the inverse function, we switch the roles of x and y and solve for y. So, we have:
x = (y - 2)^2.
To isolate y, we take the square root of both sides:
√x = y - 2.
Then, by adding 2 to both sides, we obtain:
y = √x + 2.
Now that we have the inverse function y = f^(-1)(x) = √x + 2, we can find its derivative with respect to x.
We can differentiate y = √x + 2 using the power rule for differentiation. The power rule states that the derivative of x^n, where n is a constant, is n*x^(n-1).
The derivative of √x with respect to x is (1/2)*x^(-1/2), and the derivative of 2 with respect to x is 0. Therefore, the derivative of y = f^(-1)(x) = √x + 2 is y' = (1/2)x^(-1/2).
Regarding your second concern about the implicit solution, when we solve the equation x = y^2 - 4y + 4 for dy/dx implicitly, we need to differentiate both sides of the equation with respect to x, taking into account the chain rule.
Differentiating x = y^2 - 4y + 4 implicitly, we have:
1 = 2y(dy/dx) - 4(dy/dx).
Rearranging the terms, we get:
dy/dx = 1 / (2y - 4).
So, the correct answer is "where x and y satisfy the equation y = x^2 - 4x + 4," not "where x and y satisfy the equation x = y^2 - 4y + 4."
In summary, the derivative of y = √x + 2, which is the inverse function of y = x^2 - 4x + 4 on the interval where the inverse function exists, is y' = (1/2)x^(-1/2).