![\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:
can you put the number line
Step-by-step explanation:
Answer:
Here c represents the total number of children in choir.
As per the statement:
Total number of boys in choir = 12
It is also given that: Three sevenths of a children's choir are boys.
⇒ Total number of boys = ![\frac{3}{7} c](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B7%7D%20c)
or
![12 = \frac{3}{7}c](https://tex.z-dn.net/?f=12%20%3D%20%5Cfrac%7B3%7D%7B7%7Dc)
By cross multiply we have;
84 = 3c
Divide both sides by 3 we have;
c = 28
Therefore, the total number of children in choir is 28 children
(9x + 5) - (4x+3) = 0
9x - 4x + 5 - 3 = 0
5x + 2 = 0
5x = -2
5x/5 = -2/5
x= -2/5
Answer:
120
Step-by-step explanation:
area = 15 x 8