# 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 apprecitated....thanks in advance

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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)**

switch the y and x to make the equation. x=y^2-4y-1. then solve for "Y". it will give you theinverse of the equation. and the domain can be found by setting "y cannot equal 0". so make the equation equal 0. "0=inverse function" you might have 2 answers with it being a squared equation. it depends on the inverse