Given the function 'f(x)=x^2-4x-1' find the inverse of the function f, stating its domainA break down of how the answer is got would be much...

Given the function f(x) = x^2 - 4x
-1

We need to find the inverse function of
f(x).

Let us assume that y=
f(x).

==> y = x^2 - 4x
-1

We will complete the square
.

==> y = x^2 - 4x -1 +4
-4

==> y= x^2 -4x +4 -
5

==> y = (x-2)^2 -
5

Now we will add 5 to both
sides.

==> y+5 =
(x-2)^2

Now we will take the square root of both
sides.

==> sqrt(y+5) =
x-2

Now we will add 2 to both
sides.

==> x = sqrt(y+5) +
2

Then the inverse function
is:

f^-1 (x) = sqrt(x+5) +
2

Now the domain are x values such that the
function is defined.

We know that sqrt(x+5) is defined when
sqrt(x+5) >= 0

==> sqrt(x+5) >=
0

==> x+5 >=
0

==> x >=
-5

Then, the domain is x = [-5,
inf)