Find the inverse function?
931612
f^{-1}(y) = (y-42) / 7\ns
986272
f(x) = 3(2/x + 1)\ng(x) = x^2\n\n\nHope this helps.
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
666213
the answer is linear function:\nf(n)=2n-1
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]
999317
(1) Simple inverse is log X as long as log is defined for the specific value of X, except that it's also not defined at X=1 since log(1) = 0.\n(2) Inverse function: If Y = 1/logX, then X = exp(1/Y) assuming natural logarithms. This is defined except at Y=0, which can't hold for finite X.\n\nThe proofs of both these are trivial. They are obvious.
437876
Take the derivative:\n\nf'(x) = 2x - 4\n\nSet the derivative equal to 0, and solve for x:\n\n0 = 2x - 4\nx = 2\n\nSub this value of x into the original equation to find the turning point:\n\nf(2) = (2)^2 - 4(2) + 1\nf(2) = -3\n\nSo, your turning point is (2, -3).\n\nAnd, since the x-squared term in the equation is positive, you know that this point is a minimum.\n\nDone and done.
696517
Y= 5x+3\ny-3=5x\n(y-3)/5= x\nso the inverse function is:\nf^{-1}(y)= (y-3)/5\n\nx
1013129
Please explain what is ~f.  I have curiosity, because neither inverse multiplicative, complex conjugate or inverse function could do that result...\n\nIt should be interesting!! Please post what is ~f !
492321
The inverse function of f(x) = e^x is ln(x) because \nln(e^x) = x.  \nWhen x = 1, ln(e^1)=1 ; When x = e, ln(e^e) = e\nThe equation in exponential form is:\ne^ln(e^x) = e^x\n\nIf you have a function g(x) which is the inverse function of f(x), then g(f(x)) = x. For example, arcsine(x) is the inverse function sine(x) and arcsine(sine(x))= x
669001
e^f(x) = x is the best answer i can give you.
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
195858
Here is an easy way. Say you want to find the inverse of the 2x2 matrix\n\n4 1\n1 2\n\nTo type this matrix into a TI-89, you can type in a list of lists using the square brackets, like this:\n\n[[4,1],[1,2]]\n\nYou can also use semi colons to separate the rows instead of using sublists, which can be a bit easier to read, like this:\n\n[4,1;1,2]\n\nNow to find the inverse, just type ^(-1) onto the end of the matrix you typed in (and make sure you use the unary minus key, which is the gray (-) key on the number pad, not the black subtraction key which is on the right side of the calculator)\n\nAll put together, to find the inverse of the matrix above, you can type:\n\n[4,1;1,2]^(-1)\n\nand then push enter of course.\n\nAs another example, to find the inverse of the 3x3 matrix\n\n1 2 3\n5 4 8\n9 6 7\n\ntype in:\n\n[1,2,3;5,4,8;9,6,7]^(-1)
874355
Using matices to do this\n\n[ 1 -2 1] * [x] = [-5]\n[ 2 -1 1]   [y]    [ 1]\n[2  1  2]   [z]    [10]\n\nAX = B\nA*A^(-1)X = A^(-1)B\nX = A^(-1)B\n\nSO we need the inverse of A this can be worked out using a long and tedious method or just use a calculator\n\nA^-1 = [ -.6 1 -.2]\n           [ -.4 0   2]   \n           [.8   -1 .6]\n\nA^-1B = [2}\n              [4}\n              [1]\nso x =2  y=4 z=1\n\n now you do the other one\nCheck\n2 - (4*2) +1 = 2 - 8 +1 = -5
801012
inverse square root of a number say"x" is\nx^[-(1/2)]
490567
=-1\n2(0)+3(0)-1=f(x)\n2(0)=0\n3(0)=0\nminus 1\nequals negative 1 or -1
119382
y=log_4 X\nand\ny= -log_3 (X+s) + 2\n\nRemember that logarithmic function y = log_b x is the inverse of the exponential function y = b^x, so you just graph its inverse (the exp function) and the graph of your log function is its mirror image with respect to the line \ny = x (The line that passes through (0,0) and has a slope of 1)\n\nIn your first given,\ny = log_4 x\nIts inverse function is\ny = 4^x\n\nNow graph y = 4^x\n\nNext, graph y = x and use this Line as your guide mirror to draw the log graph. This means that the shape of the log graph is the exact mirror image of the exp graph.\n\nIn your second given\ny= -log_3 (x + s) + 2\nJust rearrange the terms\ny - 2 = -log_3 (x + s)\n-(y - 2) = log_3 (x + s)\n\nThe inverse is\n-(y - 2) = 3^(x + s)\nGraph this one, then use y = x again as your "mirror."  Simple.\n\nMay this work of mine help you in your studies ;-)