Question

question 13
1 pts
which best expresses the inverse of $f(x)=(x+1)^2-2$
$\bigcirc f^{-1}(x)=\sqrt{x-2}-1$
$\bigcirc f^{-1}(x)=(x-1)^2+2$
$\bigcirc f^{-1}(x)=\sqrt{x-2}+1$
$\bigcirc f^{-1}(x)=\sqrt{x+2}-1$

Explanation:

Step1: Set $y=f(x)$

$y=(x+1)^2 - 2$

Step2: Swap $x$ and $y$

$x=(y+1)^2 - 2$

Step3: Isolate the squared term

$x+2=(y+1)^2$

Step4: Solve for $y$

Take square root of both sides: $\sqrt{x+2}=y+1$, then rearrange: $y=\sqrt{x+2}-1$

Answer:

$f^{-1}(x)=\sqrt{x+2}-1$ (the last option)