# Given the function 'f(x)=x^2-4x-1'  find the inverse of the function f, stating its domain A break down of how the answer is got would be much apprecitated....thanks in advance 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...

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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)

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