How do you solve f(x+1) if f(x) = (6x^3)-(3x^2)+(4x)-(9)?
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
1059114
3x^2 - 9x + 6 - 3(x - 1)(x - 2)\n3x^2 - 9x + 6 - 3(x^2 - 2x - x + 2)\n3x^2 - 9x + 6 - 3(x^2 - 3x + 2)\n3x^2 - 9x + 6 - 3x^2 + 9x - 6\n\nEverything gets cancelled out, so it doesn't matter if i put 10 in for x or not.\n\n3(10)^2 - 9(10) + 6\n3(100) - 90 + 6\n300 - 90 + 6\n210 + 6\n216
926616
2x+3<6x-1\n2x-6x<-1-3 \n-4x<-4\nx>1
256079
x(x^2 -9)= x (x-3)(x+3)
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.
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
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?
1167006
If the zeros are: -4, 0, and 2, our factored equation must look like this:\nf(x) = x(x+4)(x-2)\n\nFoil it out:\n\nf(x) = x(x2 + 4x - 2x - 8)\nf(x) = x(x2 + 2x -8)\n\nf(x) = x3 + 2x2 -8x\n\nAnswer number 3.\n\nHope that helps.
572480
breaking it apart, we have:\n\n(x+1)^3=(x+1)(x+1)(x+1)=\n(x^2 + 2x + 1)(x+1) =\n(x^3 + 3x^2 + 3x + 1) SO\n6(x+1)^3 = 6(x^3 + 3x^2 + 3x + 1) = 6x^3 + 18x^2 + 18x + 6\n\n(x+1)^2 = (x+1)(x+1)=\nx^2 + 2x + 1   SO\n-3(x+1)^2 = -3(x^2 + 2x + 1)=\n-3x^2 - 6x - 3\n\n4(x + 1) = 4x + 4\nand -9 stays the same.\n\nSo putting everything together, you have\n(6x^3 + 18x^2 + 18x + 6) + (-3x^2 - 6x - 3) + (4x + 4) - 9 \ncombine like terms and you get:\n6x^3 + (18x^2 -3x^2) + (18x - 6x + 4x) + (6 - 3 + 4 - 9)\n6x^3 + 15x^2 +  16x - 2\n\nWHEW!  That took me back to college pre-calc!
741149
1.f(x)=x^2+4x-7\nx=[-4+/-rt(16+28)]/2\n=[-4+/-2rt(11)]/2\n=-2+/-rt11\n\n2.3x^2-11x-20=0\n=3x^2-15x+4x-20=0\n=(3x+4)(x-5)=0\nx=-4/3 or 5
911592
Factorize!\n4x (x^2-9) = 0\n4x (x+3) (x-3)\n\nyour solution is \nX = 0\nx= -3\nx= 3
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
791940
In order to a od function the function have to obay:\nf(x) = - f (-x) <=>\n-3x^9 + 4x^5 = -(-3(-x)^9 + 4(-x)^5) <=>\n-3x^9 + 4x^5 = 3(-x)^9 - 4(-x)^5 <=>\n-3x^9 + 4x^5 = -3x^9 + 4x^5 ( True)\nthe function is od.
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
191567
Please use parentheses to explain the problem more effectively. Assuming the problem as,\n\n(x-1)/2x + x/(x+3)\n\nThe solution can be,\n\n(3x^2+2x-3)/(2x^2+6)
243121
first square two sides\nyou will get 3x+1 =9+6√(x-4)+x-4\n= (x-2)=6√(x-4)  square it \nyou will get x^2 -4x +4=9(x-4)\n=x^2 -13x + 40=0\nfactor  (x-5)(x-8)\nso x=5 or x=8