It’s 15 for #16 and then for #18 it would be 55 then for 20 it would be 6
I'm guessing you're given the function 
, and you're asked to find the inverse function 
. To do this, swap 
and 
, then solve for :
![x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}](https://tex.z-dn.net/?f=x%3D2-y%5E3%5Cimplies%20y%5E3%3D2-x%5Cimplies%20y%3D%282-x%29%5E%7B1%2F3%7D%3D%5Csqrt%5B3%5D%7B2-x%7D)
so that the inverse function is
![y^{-1}(x)=\sqrt[3]{2-x}](https://tex.z-dn.net/?f=y%5E%7B-1%7D%28x%29%3D%5Csqrt%5B3%5D%7B2-x%7D)
Just to verify:
![y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x](https://tex.z-dn.net/?f=y%28y%5E%7B-1%7D%28x%29%29%3Dy%28%5Csqrt%5B3%5D%7B2-x%7D%29%3D2-%28%5Csqrt%5B3%5D%7B2-x%7D%29%5E3%3D2-%282-x%29%3Dx)
![y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x](https://tex.z-dn.net/?f=y%5E%7B-1%7D%28y%28x%29%29%3Dy%5E%7B-1%7D%282-x%5E3%29%3D%5Csqrt%5B3%5D%7B2-%282-x%5E3%29%7D%3D%5Csqrt%5B3%5D%7Bx%5E3%7D%3Dx)
But in case you're actually only interested in computing the square root, first we note that 
(the real-valued square root) is only defined as long as 
. So 
is defined as long as 
, or 
, or equivalently 
. Under this condition, we could write
We can simplify this further, but we have to be careful. Suppose 
. Then 
. But we get the same result if 
, since 
. There are two possible values of that given the same value of 
, so to capture both of them, we take 
, the absolute value of . Then
We can't simplify the square root term further than this.
Answer:
b
Step-by-step explanation:
Answer:
Thank you and please rate me as brainliest as it will help me to level up
Answer:
C. (-3, 1)
Step-by-step explanation:
