Given f(x) = x2 + 3x – 1, find f(0).?
490567
=-1\n2(0)+3(0)-1=f(x)\n2(0)=0\n3(0)=0\nminus 1\nequals negative 1 or -1
559365
f(x)=x+3\nif x=0, f(0)=0+3=3\nif x=2, f(2)=2+3=5\nif x=-2, f(-2)=-2+3=1\n\nfor g(x)=2x+1\nif x=0, g(0)=2*0+1=1\nif x=2, g(2)=2*2+1=5\nif x=5, g(5)=2*5+1=11\n\nfor h(x.y)=2x+y\nif (x,y)=(0,0) h(0,0)=2*0+0=0\nif (x,y)=(2,1) h(2,1)=2*2+1=5\nif (x,y)=(1,3) h(1,3)=2*1+3=5
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
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
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.
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.
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
842647
Take the antiderivative (indefinite integral) of little f. This will give you some function of x plus a constant. Then plug zero into the antiderivative, set the whole expression equal to zero (to satisfy F(0)=0) and solve for the constant. Now that you know F(x). Plug in three. This is the answer.
735529
f(x)=-2x^4+6x+7\nf(-x)=-2(-x)^4+6(-x)+7\n=-2x^4-6x+7\nnether odd nor even
268383
2x^2+6x-1x-3\n2x(1x+3)-1(1x+3)=0\n(2x-1)(x+3)=0\nx=1/2 or -3
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
220918
2*x^2 - 3*x - 2 = 0 <=> x = -1/2 V x = 2 \n\n==> 2*x^2 - 3*x - 2 <=0 <=> x = [-1/2 , 2]
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
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
567979
inverse of f(x) is (x-1)/2\n\nworking:\nlet inverse of f(x) be y\n       2x+1  = y\n       2x    = y-1\n        x    = (y-1)/2\nthus, inverse of f(x)= (x-1)/2
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7