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Which function is the inverse of
[tex]f(x)=x^2-16[/tex]
if the domain of f(x) is x ≥ 0?
The inverse funtion of the given function is given by
[tex]\mathbf{f(x)=\sqrt{x+16}}[/tex]
where the domain is x ≥ -16.
Given the function is,
[tex]f(x) =x^2-16[/tex]
Let, y = f(x)
[tex]y = x^2-16[/tex]
Adding 16 to both sides we get,
[tex]y + 16 = {x}^{2} - 16 + 16[/tex]
[tex] {x}^{2} = y + 16[/tex]
Now, square rooting to both sides we get, since x ≥ 0
[tex]\sqrt{x^2}=\sqrt{y+16}[/tex]
[tex]x=\sqrt{y+16}[/tex]
[tex]f^{-1}(y)=\sqrt{y+16}[/tex]
Hence substituting x in place of y,
[tex]f^{-1}(x)=\sqrt{x+16}[/tex]
Above function is the required inverse function where the domain of that is given by, x ≥ -16.
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