The function ƒ(x) = −(x + 3)2 − 4 is not one-to-one. Find a portion of the domain where the function is one-to-one and find an inverse function.

The restricted domain for ƒ is:
a. [-3-,∞) and F^-1(x)=-3-√x+4
b.[-3-,∞) and F^-1(x)=-3+√x+4
c.[-3-,∞) and F^-1(x)=-3-√-x-4
d.[-3-,∞) and F^-1(x)=-3+√-x-4

Question

The function ƒ(x) = −(x + 3)2 − 4 is not one-to-one. Find a portion of the domain where the function is one-to-one and find an inverse function.
 	
	The restricted domain for ƒ is:
a. [-3-,∞) and F^-1(x)=-3-√x+4
b.[-3-,∞) and F^-1(x)=-3+√x+4
c.[-3-,∞) and F^-1(x)=-3-√-x-4
d.[-3-,∞) and F^-1(x)=-3+√-x-4

misty1212 · Answer

HI!!

anonymous · Answer

Hello

misty1212 · Answer

all of your domains are the same, so no worries there
it is only a matter of finding the inverse right?

anonymous · Answer

yes

misty1212 · Answer

ok lets solve 
$$x=-(y+3)^2-4$$ for $y$

misty1212 · Answer

add four get 
$$x+4=-(y-3)^2$$

misty1212 · Answer

change the sign of both sides get 
$$-x-4=(y-3)^2$$

misty1212 · Answer

take the square root of both sides get 
$$\sqrt{-x-4}=y-3$$ 
normally it would be $\pm\sqrt{-x-4}$ but we have restricted the domain so only the plus

anonymous · Answer

oh ok I get it

misty1212 · Answer

i made a typo

misty1212 · Answer

$$\sqrt{-x-4}=y+3$$

anonymous · Answer

thats how the 3 becomes negative

misty1212 · Answer

yeah but let me check something else

anonymous · Answer

ok