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
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hala718
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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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