![\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
Solution of (6x + 2.4): 3 ⁵/₉ = 2.25 : ¹/₃ is 3.6
Further explanation
Order of Operations in Mathematics follow the following rule :
Parentheses
Exponents
Multiplication and Division
Addition and Subtraction
This rule is known as the PEMDAS method.
In working on a mathematical problem, we first calculate an operation that is in parentheses, followed by exponentiation, then multiplication or division, and finally addition or subtraction.
The linear equation (y = mx + b) for this particular line is d = 0.6t
it's the same as finding the equation for any line except they use d instead of y and t instead of x. The slope m is 0.6 and intercept is zero.
Answer:
F. 300,000
Step-by-step explanation:
logically, it is decreasing by 1.
