f(x) = (x-1)(x-3) So, the vertex is at x=2. The restriction on the domain ensures that we only have the left branch of the parabola, so f is 1-to-1 and will have an inverse.
x = y^2-4y+3
x = (y-2)^2 - 1
x+1 = (y-2)^2
y-2 = ±√(x+1)
we only want the left branch, so
y-2 = -√(x+1)
y = 2-√(x+1)
Note that the range of the inverse is y<=2, the domain of the original function. The domain of the inverse is x >= -1, which is the range of the original f(x).
Find inverse of f if f(x)= x^2-4x+3, (for x is smaller than and equal to 2).
First prove that f(x) is one to one in the defined domain of f and then obtain the inverse function.
I know how to find the inverse. We just switch x and y.
so y=x^2-4x+3 becomes
x=y^2-4y+3
Inverse of f= y^2-4y+3
How do I prove that it is one to one?
I know we prove by making a= b but I'm very confused with this problem
1 answer
1 answer