the direction was to find f^(-1)x.
f(x) = x^2+2x+1, x >= -1
as a final answer, I got:
f^(-1)x = sqrt(x) - 1, x >= 0
is this right? I don't understand the restriction rules.
It is right. If originally x>=-1, that puts f(x) >=0, and in the inverse function, this restriction travels to the new domain.
kristie, Bob is correct. The restriction rule carries to f-inv.
In general, even funcions, those with a poly. whose degree is an even integer, must be restricted. They are not 1-1. 1-1 means that each horizontal line through f(x) intersects f in at most one point.
Note that restricting f here makes it 1-1, thus invertible.
To find the inverse of the function f(x) = x^2 + 2x + 1, there are a few steps you can follow:
1. Start with the equation f(x) = y.
2. Swap the x and y variables, so that you have x = y^2 + 2y + 1.
3. Rearrange the equation to isolate y. In this case, you need to complete the square:
x = (y + 1)^2
Square root both sides of the equation:
sqrt(x) = y + 1
4. Solve for y by subtracting 1 from both sides:
sqrt(x) - 1 = y
Now that you have the equation y = sqrt(x) - 1, this represents the inverse function f^(-1)(x). The restriction x >= -1 remains, as it is a result of the original function's domain.
However, it is important to note that when finding the inverse of a function, you also need to consider the range of the original function. In this case, the original function f(x) = x^2 + 2x + 1 has a range of y >= 0, which carries over to the inverse function. Therefore, the correct final answer is:
f^(-1)(x) = sqrt(x) - 1, x >= 0
So, your final answer is indeed correct.