![\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:
Thats false
Step-by-step explanation:
When you multiply by a power your basically multiplying that number b itself 10x except for 1
Answer:
114.666666667
Step-by-step explanation:
Step 1:
6/86 = 8/x Equation
Step 2:
6x = 688 Multiply
Step 3:
x = 688 ÷ 6 Divide
Answer:
x = 114.666666667
Hope This Helps :)
By "which is an identity" they just mean "which trigonometric equation is true?"
What you have to do is take one of these and sort it out to an identity you know is true, or...
*FYI: You can always test identites like this:
Use the short angle of a 3-4-5 triangle, which would have these trig ratios:
sinx = 3/5 cscx = 5/3
cosx = 4/5 secx = 5/4
tanx = 4/3 cotx = 3/4
Then just plug them in and see if it works. If it doesn't, it can't be an identity!
Let's start with c, just because it seems obvious.
The Pythagorean identity states that sin²x + cos²x = 1, so this same statement with a minus is obviously not true.
Next would be d. csc²x + cot²x = 1 is not true because of a similar Pythagorean identity 1 + cot²x = csc²x. (if you need help remembering these identites, do yourslef a favor and search up the Magic Hexagon.)
Next is b. Here we have (cscx + cotx)² = 1. Let's take the square root of each side...cscx + cotx = 1. Now you should be able to see why this can't work as a Pythagorean Identity. There's always that test we can do for verification...5/3 + 3/4 ≠ 1, nor is (5/3 + 3/4)².
By process of elimination, a must be true. You can test w/ our example ratios:
sin²xsec²x+1 = tan²xcsc²x
(3/5)²(5/4)²+1 = (4/5)²(5/3)²
(9/25)(25/16)+1 = (16/25)(25/9)
(225/400)+1 = (400/225)
(9/16)+1 = (16/9)
(81/144)+1 = (256/144)
(81/144)+(144/144) = (256/144)
(256/144) = (256/144)
