Respuesta :

ANSWER:

[tex]\:D.\text{ }f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge-4[/tex]

STEP-BY-STEP EXPLANATION:

We have the following equation:

[tex]f(x)=\sqrt{x}-4[/tex]

The inverse is the following (we calculate it by replacing f(x) by x and x by f(x)):

[tex]\begin{gathered} x=\sqrt{f^{-1}(x)}-4 \\ \\ \sqrt{f^{-1}(x)}=x+4 \\ \\ f^{-1}(x)=(x+4)^2 \end{gathered}[/tex]

The domain would be the range of the original equation, and it would be the range of values that f(x) could take, which was from -4 to positive infinity, that is, f(x) ≥ -4.

Therefore, the domain is x ≥ -4.

So the correct answer is D.

[tex]\:f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge -4[/tex]

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