f(x)= x^2 - 4x + 5 , x > 2 a) Express f(x) in the form (x+b)^2 + c b) State the range of f c) Find an expression for f^-1(x) and state its...

To express  f(x) in the form (x+b)^2 + c, we'll re-write
the expression of f(x) as:

x^2 - 4x + 5 = x^2 - 4x + 4 +
1

We notice that we have substituted the term 5 by the
sum 4+1. In this way, we've completed the square x^2 -
4x.

So, we could write the expression of
f(x):

f(x) = x^2 - 4x + 4 + 1 = (x-2)^2 +
1

From enunciation, we know that the domain of definition
of function is (2 , +infinite).

So, to calculate the range
of the function, we'll write:

f(x) =
y

But, f(x) = (x-2)^2 + 1

So,
y = (x-2)^2 + 1

We'll subtract 1 both
sides:

y - 1 = (x-2)^2

We'll
raise both sides to the power of (1/2):

sqrt(y - 1) = sqrt
[ (x-2)^2]

Since x>2 => sqrt [ (x-2)^2] =
(x-2)

x-2 = sqrt (y-1)

We'll
add 2 both sides:

x = sqrt (y-1) +
2

x>2 => sqrt (y-1) + 2 >
2

We'll subtract 2 both
sides:

sqrt (y-1) >
0

We'll raise to square both
sides:

y-1 > 0

y
> 1

So, the range of te function is: (1 ;
+infinite).

When we've determined the range, we've obtained
the expression of the inverse function f^-1(x).

f^-1(x) =
sqrt (x-1) + 2