# 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 domain^2 means squared and x > 2 should be 'x...

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 domain

^2 means squared

and x > 2 should be 'x is greater than or equal to 2'

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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

x^2-4x+5 = (x^2-2*4x+2^2 )-2^2+5

x^2-4x+5 = (x-2)^2 -4+5

x^2-4x+5 = (x-2)^2+ (-1) is required form for f(x).

b)

To find the range of f(x). It is given that x>2.

Therefore f(x) = (x-2)^2 - 1. f(x) is minimum for when x =2 and f(2) = (2-2)^2 -1. = -1.

Therfore range of f(x) i > -1. Or the ranfe is a value which belongs to the interval (-1 , infinity).

f(x)= x^2 - 4x + 5 , x > 2

a) Express f(x) in the form (x+b)^2 + c

Let us rewrite :

**f(x) = x^2 - 4x + 4 + 1 = (x-2)^2 + 1 **

b) State the range of f

f(x) = x^2 - 4x + 5

==> f'(x) = 2x - 4 = 0

==> x= 2

Then 2 is critical value for f, where fa has a minimun.

Then f(2) = 1

when x < 2 ==> f < 1

when x > = 2 ==> f >= 1

c) Find an expression for f^-1(x) and state its domain

Let f(x) = y = x^2 - 4x + 5

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

==> y-1 = (x-2)^2

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

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

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

The domain is x values such that f^-1 is defined.

f^-1 is NOT defined when x-1 < 0 ==> x< 1

Then the domain is x = [ 1, inf)