![\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:
Yes it is
Step-by-step explanation:
y=-9x means that for every x value, the y value decreases by 9, so it is proportional because the value of y decreases the same amount each time. For example, if x=1, y=-9. If x=5, y=-45, and so on.
Answer:
with my own opinion the answer is b
Answer:
y =3x - 5
Step-by-step explanation:
Well one of the way to do this question is the formula given below your question, but for me, i rather do it this way
OK they gave u y=3x - 12
And they say another line is parallel, once u see the word parallel means they have the same angle, same slope, same gradient ( sorry if I give u the wrong terms, since I learn mine in Malay)
So first u got to understand the slope-intercept form
y = mx + c
y is your y coordinate
x is your x coordinate
m is your gradient ( as in how much the line slants)
c is the y - intercept, as in, if u draw a graph, the y-coordinate where your line touches the y-axis (the vertical line)
So since the say the both lines are parallel, so both of their "m" is the same that is 3. Why? Cuz just compare
y = mx + c
y = 3x - 12
Therefore the m is 3
Therefore u alredy got your first value, so put it in your slope intercept form
y = mx + c
y = 3x + c
Now how to find c? Very easy, just put in the value of y and x from the coordinate ( that the line youre finding is on) they gave u, just put the value in
They gave u
(1, - 2)
So when x = 1, y = - 2
Therefore put it in
y = 3x + c
-2 = 3 (1) + c
c = - 5
Since now u have your m and c value, your slope intercept form is done, just put it in
y = mx + c
y = 3x + (-5)
y = 3x -5
(a) It is leaking 18 liters per minute.
Explanation : From 15 minutes to 20 minutes, your starting point at 15 is 590 liters and at 20 you have exactly 500. Next you have 25 which is now at 410. So it's leaking 90 liters every five minutes, so of course dividing the two (90 ÷ 5) you get 18.
(a)² it increases for about the same time that it would decrease of course.
(b) The vat started at 860 liters before they started draining it.
Explanation : Well, all it takes is just working backwards, add 90 liters to the liters at 15 minutes and deduct five minutes from 15. Do this process until you arrive at 0. It's simple
