if f(x)=1-(x-1)^1/2  find f(-x)?
153289
The answer is: f(-x)=1-(-x-1)^1/2.\n\nIf f(x)=f(x)=1-(x-1)^1/2, then it means that f(y)=1-(y-1)^1/2. Let y=(-x), then we get f(-x)=1-(-x-1)^1/2. Note that -x must be less than -1 in order to get the answer in real number. Otherwise, it will be a complex number.
491455
(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
655260
2=-1=x\n2+1=x\nthere fore x =3
127356
7 - (4-x)^1/2= 4\n-(4-x)^1/2 = 4-7\n-(4-x)^1/2 = -3\n(4-x)^1/2 = 3\n(4-x)=3^2=9\n4-x=9\n-x=9-4\n-x=5\nx=-5
491468
If by x1/2 you mean x^(1/2), and by -1-[x] you mean -1 minus the absolute value of x, then the composite function f o g is (-1-[x])^(1/2).\n\nWhat is unusual about this is that this function is not defined for any real number x.  The expression (-1-[x]) is always going to be negative, no matter what x is.  Raising this to the power of 1/2 is the same as taking the square root of it.  Since the square root of a negative number is not defined in the real number system, the domain is the empty set.
106861
How about:\nx ^2/(x^2-4)+1/(x+2) = 0\nx^2/(x^2-4) = -1/(x+2)\nx^2/(x+2)(x-2) = -1/(x+2)\nx^2/(x-2) = -1\nx^2 = -x + 2\nx^2 + x – 2 = 0\n(x+2)(x-1) = 0\ncan you take it from here?
818927
if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed
965701
f(x) = x(x+2)(x-2)(x-4)\n\nMultiply it out if you need it in general form.\n\nThere could also be a constant in front of the x.
223472
x - 1 / x + 3 = x + 2 / x - 4\nx-1/x-x-2/x = -3-4\n-1/x-2/x = -7\n-3/x = -7\n7x = 3\nx=3/7
692175
( x^2 -x +2 )( x^2 + x + 2 )\n\n=>\n a= -1\n b= +2\n c= +1\n d= +2
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
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)
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
607013
Another Method.\n\n-1 = -1\n\n-1/1 = 1/-1\n\nSquare root both sides,\n\nSQR(-1/1) = SQR (1/-1)\n\nExpanding,\n\nSQR(-1)/SQR(1) = SQR(1)/SQR(-1)\n\nSQR(-1)/1 = 1/SQR(-1)\n\nCross Multiply,\n\nSQR(-1)*SQR(-1) = 1\n\n(SQR(-1))^2 = 1\n\n-1 = 1\n\nDivide by 2 on both sides,\n\n-1/2 = 1/2\n\nAdd 3/2 to both sides,\n\n-1/2 + 3/2 = 1/2 + 3/2\n\n1 = 2 (ans)
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
736337
2/x + 3 = 2\n2/x = -1\n-x = 2\nx = -2
869813
(y--1)/(-2--1)=(x-1)/(2-1)\n=(y+1)/(-2+1)=(x-1)/(1)\n=>y+1=-1(x-1)\ny+1=-x+1\ny=-x is the required equation