What happens to Y? y = f(x) = 2x^2 - 3x^2?
824521
this just means Y is a function of x\n\ny = x^2(2-3)\ny = -x^2
708317
Ok, so if I have this correctly ... f(x) = 3x + 7 and you want to figure out the value of 2f(x²) - 5.\n\nSo first, you have to substitute x² into your equation for f(x) to obtain:\n3x² + 7.\n\nNow, this is the value for f(x²), so we can substitute it into the sexond equation:\n\n2f(x²) - 5\n=2(3x² + 7) - 5\n\nNow just expand and simplify.\n\n2(3x² + 7) - 5\n=6x² + 14 - 5\n=6x² + 9\n= 3(2x² + 3)
201432
x=2 and y =2,\n\n2^2=2^2 and 2*2 = 2 + 2 =4
717628
f(x) = -x^3 - x^2 + x - 9\nf(-1) means replace x with -1\nf(-1) = -(-1)^3 - (-1)^2 + (-1) - 9\nf(-1) = -(-1) - (1) - 1 - 9\nf(-1) = 1 - 1 - 1 - 9\nf(-1) = -10\n\nf(0) means replace x with 0\nf(0) = -(0)^3 - (0)^2 - 0 - 9\nf(0) = 0 - 0 - 0 - 9\nf(0) = -9\n\nf(2) means replace x with 2.\nTry this one on your own.
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
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.
678157
2x^2(1-2x)=0\n\nx=0\nx= (1/2)
838016
This is basically just asking what is f(x)/g(x)\n\nso your answer would be \n\n(x+2) / (3x^2 - 5)
879979
f(x)=(x^2-2x)(x^2+2.44x+1.4884)\n =(x^4+2.44x^3+1.4884x^2-2x^3-4.88x^2- 2.9768x)\n =(x^4 + 0.44x^3 - 3.3916x^2 - 2.9768x)\n\nF(x)=(x^5)/5+(0.44x^4)/4-(3.3916x^3)/3-             (2.9768x^2)/2+c\n\nam i really asked a Q?
1112480
Carefully, very carefully :)\n\nf(x) =-1/3x^2 is a parabola origin at (0,0) with branches downward.\n\n\nf(x)=x^2-6x+9 =(x-3)^2 orgin at x=3, y intecept at x=0
847176
f(x,y) = x^2*y*(e^(-x^2-y^2))\n\nfx(x,y) = 2x*y*e^(-x^2 - y^2)(1-2x^2) this is wrong:\nf_x(x,y) = x^2y ( e^(-x^2-y^2)) (-2x) + 2xye^(-x^2-y^2))\n= 2xye^(-x^2-y^2) [-x^2 + 1] =0\nif x=0 or y=0 or x^2 =1, \nx=0, y=0, x=1 or x=-1\n\nfy(x,y) = x^2*e^(-x^2 - y^2)(1-2y^2) \nf_y(x,y)=  x^2 y (e^(-x^2-y^2)) (-2y) +  x^2(e^(-x^2-y^2))\n=x^2e^(-x^2-y^2) [-2y^2 + 1] \nx=0, y=1/sqrt(2),y=-1/sqrt(2)\n\nso the critical points are: (0,y)\n(0,0), \n(1,1/sqrt(2)), (1,-1/sqrt(2))\n(-1,1/sqrt(2)),(-1,-1/sqrt(2))\ns
1076805
y = -3 / (4x + 2)\n\nDomain: all acceptable x values:  We know that (4x + 2), or y would equal   -3/0, and that cannot be.  So actually set (4x + 2) =0 to find out what x CANNOT be\n\n4x + 2 = 0\n4x = -2\nx = -.5\n\nDomain (-infinity,-.5)U(-.5,+infinity)\n\nRange: All acceptable y values.  Looking at a graph of the function, you can see that y cannot equal zero, and looking at the equation,  there is nothing you can divide "-3" by and get zero, so \nRange (-infinity,0)U(0,+infinity)\n\nCheck this out to verify:\nhttp://www.coolmath.com/graphit/index.html
965203
y = (ln x²)^(2x+3) = e^((2x+3) ln ln x²)\nNow just ruthlessly apply the chain rule:\ndy/dx = e^((2x+3) ln ln x²) * (2 ln ln x² + (2x+3) * 1/ln x² * 1/x² * 2x)\nAnd simplify:\n(2 ln x)^(2x+3) * (2 ln (2 ln x) + (2x+3)/(x ln x))
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
821547
f(x)=4x^2 + 9x - 2\n\nhere, this means the function (f) of x=4x^2 + 9x - 2\n\nso if there was f(2)=4x^2 + 9x - 2  that would mean solve the function if x=2
1026879
f(x,y)= (1-x)^2 + 105 (y- x^2)^2\n\nIt's often used to test out various nonlinear optimization algorithms.
106013
y = -2x + 8\n\nsolve it yourself