3 years ago

Please help pic below URGENT

Mathematics

1 answer:

Artyom0805 [142]3 years ago

5 0

There is not picture................... Is this a joke?

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ANSWER:

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

STEP-BY-STEP EXPLANATION:

We have the following equation:

$f(x)=\sqrt{x}-4$

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

$\begin{gathered} x=\sqrt{f^{-1}(x)}-4 \\ \\ \sqrt{f^{-1}(x)}=x+4 \\ \\ f^{-1}(x)=(x+4)^2 \end{gathered}$

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.

$\:f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge -4$

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