List all of the zeros of the function f(x)=x^3-4x^2+14x-20?
958467
Replace x with 2:\nf(2)=2^3-4(2)^2+14(2)-20\nf(2)=8-16+28-20\nf(2)=-8+8\nf(2)=0
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
922587
ok\nf(x) = 20 - x\nf(0) = 20 - 0 = 20\nf(2) = 20 - 2 = 18\nf(4) = 20 - 4 = 16\nf(8) = 20 - 8 = 12
547775
f(x)= x^3+x^2-22x-40\nf(x)= (x+4)(ax^2 + bx + c)\n\nComparing coefficients,\nx^3: 1=a\nconstant: -40=4c => c=-10\nx^2: 1=b+4a => 1=b+4 => b=-3\n\nf(x) = (x+4)(x^2-3x-10) = (x+4)(x+2)(x-5)\n\nThe other factors are (x+2) and (x-5).
643397
(8y^3 + 27)/(2xy - 10y + 3x - 15)\n((2y + 3)(4y^2 - 6y + 9))/((2xy - 10y) + (3x - 15))\n((2y + 3)(4y^2 - 6y + 9))/(2y(x - 5) + 3(x - 5))\n((2y + 3)(4y^2 - 6y + 9))/((2y + 3)(x - 5))\nthe (2y + 3)s cancel out\n(4y^2 - 6y + 9)/(x - 5)\n\nANS : (4y^2 - 6y + 9)/(x - 5)
1149923
(2,-4) if f(x) is defined for x=2
110508
(x^2+5) / (x^3+15-2x^2+1)^2\non [0,1]
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
490567
=-1\n2(0)+3(0)-1=f(x)\n2(0)=0\n3(0)=0\nminus 1\nequals negative 1 or -1
707641
another zero would be the conjugate of the one you know,\nthen try the rational zeros test, or even syn div
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.
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
665603
you put (x+2) in for x in the equation first\n\n((x+2)^3)+(2(x+2)^2)-4\n\nthen simplify \n\nx^3 + 8x^2 + 20x + 12\n\nand multiply the entire thing by 3\n\n3(x^3 + 8x^2 + 20x + 12)\n\n3x^3 + 24x^2 + 60x + 36
860777
f'(x) = 0 to find critical #s.\n\nf'(x) = 4x^3 - 16x = 0\n16x = 4x^3\n16 = 4x^2\n4 = x^2\nx = (+ or - ) 2  therefore critical points occur when x = 2 or x = -2\n\nf''(x) = 12x^2 - 16\nwhen x = 2, f''(2) =32 > 0, and when x = -2, f''(-2) = 32 > 0, therefore F has relative minimums at x = 2, and x = -2 \n\nSince f''(x) > 0 at both points, our function "opens up" and so these values are at a minimum.
700407
f(x)=x^5-13x^3+36x \nf(x)=x(x^4 - 13x^2 + 36)\nf(x)=x(x^4 - 9x^2 - 4x^2 + 36)\nf(x)=x(x^2(x^2 - 9) - 4( x^2 - 9))\nf(x)=x(x^2 - 9) (x^2 - 4)\nf(x)=x(x-3)(x+3)(x-2)(x+2)\n\nas u havent mentioned domain or range of the function,,\nit is assumed ,,that domain is any real number,,\nthus,,the roots of ' x ' can be,,3,,-3,,2,,-2,,or any real number,,\n\nfoe example--\nf(1)=1(1-3)(1+3)(1-2)(1+2)\nf(1)=(-2) x (-4) x (-1) x (3)\nsoo,,\nf(1)=-24\n\nand ya,,f(x),,need not be zero always,,\nthis is what ppl r used to,,and soo,,they forgt that it is not zero always,,\nsoo,,\nzeros of the function here,,r,, any real number,,