determine the equation for the inverse function...?
453667
y=(x+8)^3-5\ny+5=(x+8)^3\n(y+5)^(1/3)=x+8\nx=(y+5)^(1/3)-8
887482
y=arccos(x - 5) \nDomain is 0 ≤ (x - 5) ≤ π      ie 5 ≤ x ≤ π + 5\nRange -1 ≤ y ≤ 1\n\nSo inverse function is\n\nx = arccos(y - 5)\n\nRange is 5 ≤ y ≤ π + 5\nDomain is -1 ≤ x ≤ 1\n\nie y - 5 = cosx \nSo y = cosx + 5\n\nDomain is -1 ≤ x ≤ 1\nRange is 5 ≤ y ≤ π + 5
895922
That equation is just the following when written as a matrix (// = new line)\n[ 2.7 -3.5 // 6.3 4][x // y] = [13.35 // 12.9]\nNow, you need to find the inverse of the left hand matrix. The inverse of [a b // c d] is 1/(ad-bc) * [d -b // -c a].\nSo, the inverse is 1/32.85 * [4 3.5 // -6.3 2.7].\nSo, multiplying both sides by that:\n[x // y ] = 1/32.85 * [4 3.5 // -6.3 2.7][13.35 // 12.9].\nMultiplying that out gives [x // y] = [3 // -1.5]
304526
let a-x=u\nso, -dx=du\n\nthen ∫1/(a-x)^2dx= \n-∫1/[(u)^2]du=\n-∫u^(-2)du=\n\n-[u^(-2+1)]/(-2+1)=\n\n-u^(-1)/-1=\nu^(-1)=\n\n1/u + C=\n\n1/(a-x) + C
284416
For inverse var.,\n\ny=k/x\n9=k/4, so k=36.
70526
=((-4y)^2)/(10y-5)= (16y^2)/(10y-5)\nand if the equation is equal to zero then y=4i
201836
inverse of 1/2 = 2/1 = 2\n\nunless you have X^(1/2) = sqrt(X)
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
813582
the inverse function is\n\nh =  [3(R-1)]/2\n\ntake the original function, switch the placement of the (h) and the (R) and solve for (h). \n\nat 6'0'', Reach = 7'6''\nat 6'6'', Reach = 8'3''
446101
switch the y and x\n\nx = 2y - 3\n\nThen solve for y\n\nx + 3 = 2y\n\n(x+3)/2 = y\n\nnow you have the inverse.
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2
978538
If you are going to rewrite the equation as a function of R you first need to solve for R.  You would get R = 12/I.  Now to put it into functional notation f(R) = 12/I.  To answer your other question "The inverse of a function is SOMETIMES a function".  The easiest way to explain this is to use ordered pairs.  If a function gives you ordered pairs (1,2)(3,4)(5,6) the inverse will have the ordered pairs (2,1)(4,3)(6,5).  You would just switch the x and y coordinates.  Play around with this to see that it the inverse is sometimes a function itself.  By the way in order for it to be a function you cannot have repeating y-coordinates.
604301
We just interchange the x's and the y's. then find what is y\nExample the inverse of f(x)=x+1\nSince f(x) is just equivalent to y... \nthen y=x+1. we then switch...\nx=y+1... then find for y... y=x-1. \nThat is the inverse of f(x)=x+1\nNow let's solve your problem\ny=f(x)=7x-2\ny=7x-2\nswitch x and y \nx=7y-2\nsolve for y\n(x+2)/7=y\ninverse is (x+2)/7
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
238752
ln and e^x are inverse functions. So you can use them to solve equations.\n\nA looks like it's already solved, so just use a calculator and find the ln of 0.0093, which is -4.67774\n\nFor equation B, take the ln of both sides.\n\nln(e^x) = ln(62.5)\n\nThe ln(e^x) cancels each other out.\n\nx = ln(62.5)\n\nx = 4.13517
860831
I am guessing your function to be one of the following:\n(1) f(x) = ½x - 3 or (2) f(x) = -½x -3\nIn function (1):\nf(0) = ½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = ½(2) - 3 = 1 - 3 = -2 ... or... (2, -2)\n\nIn function (2):\nf(0) = -½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = -½(2) - 3 = (-1) - 3 = -4 ... or ... (2, -4)
380912
When dealing with inverse variation there are two different equations you can use, y= k/x or first y/new y= new x/ first x. So in this case you can either do I= k/d^2 then do I= k/2d^2 and compare the I's. Another way you can do it is I original/ I new= 2d^2/ d^2 then you compare the first I and the second I. Unfortunetly in this case you dont have numbers so you just look at the fact that in the firt equation your dividing by the d^2 or the 2d^2 and you look that in the first one your just dividing by d^2 or 12 inches, while in the second your dividing by 2d^2 or 4d or 24 inches, which is more. Then when you divide by more than the I is smaller or lest intense.