yea just put a line over the 0 to indicate it's infinite
A delightful question.
I think I have it.
The number is 22 .
Its factors are 1, 2, 11, and 22 .
It would be C.60 If you add the column of 16-20 and 21-25 it makes 60.
![\bf f(x)=y=2x+sin(x) \\\\\\ inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x) \\\\\\ \textit{now, the "y" in the inverse, is really just g(x)} \\\\\\ \textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\ -----------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dy%3D2x%2Bsin%28x%29%0A%5C%5C%5C%5C%5C%5C%0Ainverse%5Cimplies%20x%3D2y%2Bsin%28y%29%5Cleftarrow%20f%5E%7B-1%7D%28x%29%5Cleftarrow%20g%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20the%20%22y%22%20in%20the%20inverse%2C%20is%20really%20just%20g%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20we%20can%20write%20it%20as%20%7Dx%3D2g%28x%29%2Bsin%5Bg%28x%29%5D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C)
![\bf \textit{let's use implicit differentiation}\\\\ 1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor} \\\\\\ 1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Blet%27s%20use%20implicit%20differentiation%7D%5C%5C%5C%5C%0A1%3D2%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%2Bcos%5Bg%28x%29%5D%5Ccdot%20%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5Cimpliedby%20%5Ctextit%7Bcommon%20factor%7D%0A%5C%5C%5C%5C%5C%5C%0A1%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5B2%2Bcos%5Bg%28x%29%5D%5D%5Cimplies%20%5Ccfrac%7B1%7D%7B%5B2%2Bcos%5Bg%28x%29%5D%5D%7D%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%3Dg%27%28x%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D)
now, if we just knew what g(2) is, we'd be golden, however, we dunno
BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)
for inverse expressions, the domain and range is the same as the original, just switched over
so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2
so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2
thus 2 = 2x+sin(x)
![\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2 \\\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}](https://tex.z-dn.net/?f=%5Cbf%202%3D2x%2Bsin%28x%29%5Cimplies%200%3D2x%2Bsin%28x%29-2%0A%5C%5C%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D%5Cimplies%20g%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5B2x%2Bsin%28x%29-2%5D%7D)
hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
Answer:
inverse
Step-by-step explanation:
¬ is the symbol for not. When we have not a implies not b we have a inverse