find the inverse function of f(x)=-2/3x+5?
373083
Well, I'm not going to answer the question for you, but I'll tell you how to do it.\n\nWhen you're graphing the function, you set up a table with f(x) and x as the two columns, then solve for x for any set of numbers, like -3, -2, -1, 0, 1, 2, 3, and that's probably more than enough. Graph those on a piece of graph paper (f(x)=y, and then x). \n\nNow, to find the INVERSE, you simple switch the X and the Y. Instead of solving for x, you solve for f(x). So, your problem would be written x= -2/3y + 5. An easier way to do it, take the table you made before, and switch the columns, so that the solutions for x are underneath f(x) and the numbers you picked for f(x) are underneath the x column. Graph those coordinates. When you look at the graph, the inverse should be a mirror reflection of the first one, with  the graph of y=x as the line of symmetry. \n\nThat's how you find the inverse of any problem. Sorry if anything's unclear, if you need a better visual, I'm sure you could google up this exact method.
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18
652392
1/2 x^2 -3x\n- 2/3 x^2 +3x -5\n________________\n\n-1/6x^2 (3x's cancel out) -5\n______________________________\n\n-1/6x^2 -5
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
623250
substitute the x in the function for the 3.  that should give you:\n\nf(3) = (-5(3)^3)+(6(3)^2)-3-4  \n\nf(3) = (-135)+(54)-7   (simplified)        \n\nand your answer is:\nf(3) = -88
894442
f(x)=2x^3-3x-1\nf(-1)=-2+3-1=0\nx+1 is a factor\ndividing by x+1\n2  0  -3  -1\n0 -2   2   1\n-------------\n2x^2-2x-1\nso the factors are\n(x+1)(2x^2-2x-1)
54462
(3x+5)/wz \n(3*4+5)/(-1*5)\n(15+5) / (-1*5) \n20/-5 = -4
943371
f(x)=a+tan(bx)\n\n(-7/2) < b < (7/2)\n\nf(0)=-5\nf(π/7)=sqrt(3)-5\n\nf(0)=-5=a+tan(0)=a\na=5\nf(-π/7)=-5+sqrt(3)=a+tan(bπ/7)\nsqrt(3)=tan(b*π/7)\narctan sqrt 3=60°=π/3\nb*π/7=π/3\nb/7=1/3\nb=7/3  so \nf(x)=-5 +tan(7x/3)
888899
f(x) = x^2 + 3x\n\nf(2-a) = (2-a)^2 + 3(2-a)\n\nExpand (2-a)^2 to (2-a)(2-a) and distribute the 3(2-a)\n\nf(2-a) = (2-a)(2-a) + 6 - 3a\n\nf(2-a) = 4 -2a -2a + a^2 + 6 - 3a\n\nThe solution is:\n\nf(2-a) = 10 - 7a + a^2
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
776562
f(g(x)) = 3(2x-7)+5\n= 6x-21+5\n=6x-16\n\ng(f(x)) = 2(3x+5)-7\n=6x+10-7\n=6x+3
149693
3 - 2x = 5\n--> -2x = 5 - 3\n--> -2x = 2\n--> x = 2/(-2)\n--> x = -1 the answer!\n\nYou can check the answer by substituting the vale in the original eqn"\n3 - 2x\n= 3 - 2 (-1)\n= 3 + 2\n= 5
921411
(x+2)(x-5)=(x-1)(x-3)\nx^2-3x-10=x^2-4x+3\n-3x-10=-4x+3 [x^2's cancel]\n+3x       +3x\n-10=-x+3\n-3        -3\n-13=-x\nx=13
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
886499
1). f(x) = (x-3)(x²+3x-5) +2. Multiply out to get your answer.\n2). f(1) =1, so we have to solve\nx³-8x²+17x-10 = 0.\nWe know x = 1 is a root of this equation, so x-1 is a\nfactor of the left-hand side. The other factor, obtained\nby long division or synthetic division is\nx²-7x+10.\nNow set this equal to 0 and solve:\n(x-5)(x-2) = 0,\nso the other 2 solutions are x = 5 and x = 2.
800627
x = 3/(x-2)\n\n=> x(x-2) = 3\n=> x^2 - 2x = 3\n=> x^2 - 2x - 3 = 0\n=> (x-3) (x+1) = 0\n\nEither x-3 = 0 => x=3\nOr x+1 = 0 => x=-1
10353
nobody can inerse this function\ninstead u can draw its graph or calculate the derivate of the inverse by the relation\n(f^-1)'(y)=1/f'(x)