Question

For the function f(x) = x^2 - 4x +5, x>= 2, which is equal to d/dx (f-1 this is the inverse)(x)) ?

1/(2y-4) where x and y are related by the equation (satisfy the equation) x=y^2-4y+5 x>= 1

2y-4 where x and y are related by the equation y= = x^2 - 4x +5, x>= 2

1/2x-4 for x>= 1

1/2x-4 for x>= 2

1/2y-4 where x and y are related by the equation y= = x^2 - 4x +5, x>= 2

inverse of x^2 - 4x +5 is 2±sqrt(x-1) and the derivative of that is (2x-4) but I don't know how to get the x>= 1 or x>= 2

Answer 1

@amistre64 @mathmate

Answer 2

@mathsciencehistory

Answer 3

@NathanJHW

Answer 4

There are 2 ways to decide between x>= 1 or x>= 2

FIrst, from your inverse function
2±sqrt(x-1)
for a real number , you want x>=1

or, from the original function
f(x) = x^2 - 4x +5, x>= 2
f(2) = 4-8+5= 1
so f(x) >= 1
and that means the inverse function's input (i.e. x ) >=1

btw, note that if we find
dy/dx = 2x-4
then
dx/dy is the derivative of the inverse function: 1/(2x-4)