one more: give f(x) = 3x +3, find f (a+1) i say a+7 is the answer......?
96099
Your anwser is incorrect.  f (a+1)=3(a+1)+3 First distribute; 3a+3+3 so the anwser is 3a+6!
1169872
(7/(3x)) / (x/3)\n(7/(3x)) * (3/x)\n21/(3x^2)\n7/x^2\n\nBut if you meant\n\n((7/3)x)/(x/3)\n((7x)/3)*(3/x)\n(21x)/(3x)\nANS : 7\n\n-------------------------------\n\n2 1/3 - 3/4\n2 1/3 = (7/3)\n\n(7/3) - (3/4)\nMultiply everything by 12\n(28 - 9)/12\n(19/12) or 1 (7/12)
763039
x(sqrt)3+4=7+(sqrt)3\n=>x(sqrt)3-(sqrt)3=7-4=3\n(sqrt)3} (x-1)=3\n3(x-1)=3(sqrt)3    [multiplying both sides by (sqrt)3]\nx-1=(sqrt)3      [cancelling or dividing both sides by 3]\nx=1+(sqrt)3  ans
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
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
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.
739947
To do this you find the first derivative:\nf'(x) = 3x^2 -8x + 3\nWhen f'(x) = 0 there is a miinimum or maximum.\nWhere is this one increasing to a max and/or min?  Set it equal to 0,\n3x^2 -8x + 3 = 0; apply the quadratic formula to find the 0's for x:\n\nx = {-(-8) +- sqr[(-8)^2 - 4(3)(3)]}/2(3)\n   = {8 +- sqr[64-36]}/2 = {8 +- sqr[28]}/2 = {8 +- 2sqr(7)}/6\n   = (4+-sqr(7))/3 = x1, x2\n\nNow you have two values of x; now look at f(x) in the interval\n(-infinity, x1), then (x1,x2), then (x2,+infinity); this will tell you the answer.  Make a table showing the values approaching x1, \nx1 to x2, then x2 on.  And there you have it.
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
779831
Answer 1:\n3(x + 1) = 24\n3x + 3 = 24\n3x + 3 - 3 = 24 - 3 (Subtract 3 from both sides)\n3x = 21\n3x/3 = 21/3 (Divide both sides by 3)\nx = 21/3\nx = 7\n\nAnswer 2:\n3(x + 2) = -9 - 2x\n3x + 6 = -9-2x\n3x + 2x + 6 = -9 - 2x + 2x (Add 2x to both sides)\n5x + 6 = -9\n5x + 6 - 6 = -9 - 6 (Subtract 6 from both sides)\n5x = -15\n5x/5 = -15/5 (Divivde both sides by 5)\nx = -15/5\nx = -3\n\nAnswer 3:\n3(x + 2) = 9 + 2(x+4)\n3x + 6 = 9 + 2x + 8\n3x + 6 = 2x + 17\n3x + 6 - 6 = 2x + 17 - 6 (Subtract 6 from both sides)\n3x = 2x + 11\n3x - 2x = 2x - 2x + 11 (Subract 2x from both sides)\nx = 11
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.
1066777
x-3=3\nx=3+3\nx=6\n\n\n15-(x)=7\n15-7=x\n8=x
884668
f(x)=-4x^3+5x^2+8\nf(-3)=-4(-3)^3+5(-3)^2+8\n=108+45+8 is not equal to zero\nso x+3 is not a factor
461122
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards
1019687
F(n)=F(n-1)+F(n-2), where F(0) = 0 and F(1) = 1\nn= 2 \nF(2) = F(2-1) + F(2-2)\nF(2) = F(1) + F(0) \nF(2) = 1 + 0 = 1\nn = 3\nF(3) = F(3-1) + F(3-2)\nF(3) = F(2) + F(1)\nF(3) = 1 + 1 \nF(3) = 2\nn = 4\nF(4) = F(4-1) + F(4-2)\nF(4) = F(3) + F(2)\nF(4) = 2 + 1\nF(4) = 3\nn = 5\nF(5) = F(4) + F(3)\nF(5) = 3 + 2\nF(5) = 5
794793
x = 3. Try entering it on your calculator.
218516
If that is supposed to be a function of x, then\nf(x) = (-2x + 7)/(4x - 3)\nSince the denominator cannot be zero, equate it with zero and you get the vertical asymptote:\n4x - 3 = 0\n4x = 3\nx = 3/4 --- Vertical asymptote.\n\nTo get the horizontal asymptote, get the inverse function.\ny = (-2x + 7)/(4x - 3)\n4xy - 3y = -2x + 7\n4xy + 2x = 3y + 7\nx = (3y + 7)/(4y + 2)\n\nThen the horizontal asymptote is the denominator equated to zero.\n4y + 2 = 0\n4y = -2\ny = -1/2 --- horizontal asymptote\n\n^_^
295299
F = first = 3x times 3x = 9x^2\nO = outer = 3x times 4 = 12x\nI = inner = -4 times 3x = -12x\nL = last = -4 times 4 = -16\n\nAdd like terms (they become 0) and write answer: 9x^2 - 16