If f(x+1) = x^2 - 3x + 2 for all real x, then f(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
915401
this means\n\nx^3 = -1\nor x^3+1 = 0\n\nf(x) = x^3+1\n\nf(-1) = 0\nso one root is -1\n\nfor the other root by division\n(x+1)(x^2-x+1)\nx^2-x+1 = 0 => x= (1+/-sqrt(3)i)/2
741552
You have got it right.\nthe answer is\n\nf(x) = 3x
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
335000
(3x-9)/(x^2-x-6)=\n\n[3(x-3)]/[x^2-3x+2x-6]=\n\n[(3(x-3)]/[x(x-3)+2(x-3)]=\n\n[3(x-3)]/[(x-3)(x+2)]=\n\n3/[x+2]
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
276641
f(x)=3x^2-2x+8 for x=6\n\nf(6)=3*(6^2)-2*6+8\nf(6)=3*(36)-12+8\nf(6)=108-12+8\nf(6)=104
1138567
ok .....\n\ni think its....\nf(g(x))\n 5(x^2 - 1) + 3\n 5x^2 - 5 + 3\n =5x^2 - 2\n                        \ni think thats it....
472136
False. For example: F(x)=x^2 increases for x>0.\n\nF(x)=x^3 always increases.
695915
Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
600301
x^3+x^2+2 = x^5+2\n\n(x^2+2)^2 = x^4+4\n\n(x^3+x^2+2)/[(x^2+2)^2] = (x^5+2)/(x^4+4) = (x+2)/4 = (x+1)/2
672142
it's\n[x(x+2)]/[x*x] -2/3x=\n\n[x+2]/x - 2/3x=\n\n[3[x+2]-2]/3x=\n\n[3x+4]/3x=\n\nx + [4/3]x
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18
881627
f(x)=x^3+5x^2+3x+1\nf'(x)=3x^2+10x+3\nsetting it to zero\n(3x+1)(x+3)=0\nx=-1/3 and x=-3\nf''(x)=6x+10\nat x=-1/3 f''(x)=+ve and so min\nat x=-3 f''(x)=-ve and so max
599364
x^2 + x = x(x+1)\n\nIt can be simplified to x(x+1).
972511
A fascinating subject.\n\nIt's recursively defined.  \n\nf(n+2) = f(n+1) + f(n) for n>0\n\nwhere f(1) = f(2) = 1.\n\nIt has some really neat properties and shows up in nature.
986272
f(x) = 3(2/x + 1)\ng(x) = x^2\n\n\nHope this helps.