How many negative real zeros does the f(x)=x^5+x^4-x^2-x+1 have ?
455928
Use Descarte's rule of signs:\n\nf(x) = x^5 + x^4 - x^2 - x + 1\n  To determine negative zeros, plug in a negative and solve.\nf(-) = (-)   (+)   (-)   (+)   (+)\n  Count the number of sign changes.\nThere are 3 sign changes, so you could have 3 or 1 negative real zeros. (The reason you could have 1, is that there might be a negative imaginary zero). You could graph it to see whether you have 3 or 1.
739088
Consider a = 1, b = 0, c = 1.  Clearly, b^2-4ac = -4 < 0.  This corresponds to the equation x^2 + 1 = 0.  But x^2 is never negative, and 1 more than that is always positive, so this has no solutions.\n\nIf you want only non-imaginaries, there are no solutions when b^2 < 4ac and only one solution when b^2 = 4ac.\n\nYou clarify: "that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions."\n\nRational equations do indeed give complex solutions, many of them nonreal.  The example I gave above (x^2 + 1 = 0) is an example.  The algebraic closure of the reals is the complex numbers; the algebraic closure of the rationals is the algebraic numbers.  If you want all quadratics to give two solutions (counting multiplicity), you'll need to allow imaginaries.
888635
Let f(x) = 3x^4 - 25 + 5/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nBecause f(x) is a polynomial function (of degree 4) on x, it follows that the function f is continuous on R (the set of real numbers).  So, if we take the limit on the left (and right) of x, we will get the same value for both sides. \n\nThat is, the behavior on the left (Left-hand Limit) will be: as x approaches  c from the left, we get f(x) approaches f(c) = 3c^4 - 25 + 5/4. Similarly, the behavior on the right (Right-hand Limit) will be: as x approaches c from the right, we also get f(x) approaches f(c) = 3c^4 - 25 + 5/4. \n\nTherefore, both the left and right-hand limit yields to f(c), i.e. as x approaches c, we always obtain f(c).\n\nHence, both left and right-hand behavior of f are the same for all real valued x.
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
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
941571
1) f(x) = A(x-1)(x-2)(x-4) for some A. Substitute in 0 to get -16 = A*-1*-2*-4, so A = 2. So 2(x-1)(x-2)(x-4).\n2) Whenever you have a polynomial with real coefficients and a+bi is a root, you also have a-bi is a root. So you know the 3 roots are 4, 3+i, 3-i. Thus f(x) = (x-4)(x-(3+i))(x-(3-i)). Multiplying those last two together, you get x^2 - x(-3-i-3+i) + (3+i)(3-i) = x^2 - 6x + (9-i^2) = x^2-6x+10. So f(x) = (x-4)(x^2-6x+10).\n3) I don't think you wrote the question down properly. Expanding the left hand side gives 3x^3 + (-a-27)x^2 + (9a+42)x - 14a. You need a=5 for the first part to work, but that doesn't give what you wrote.
110508
(x^2+5) / (x^3+15-2x^2+1)^2\non [0,1]
735726
For x = -2, f(-2) = -8 - 4 = -12\nFor x = 1, f(1) = 4 - 8 = -4\nFor x = 2, f(2) = 8 - 8/4 = 6\nFor x = 3, f(3) = 12 - 8/9 = eleven and one-ninth\nFor x = 4, f(4) = 16 - 8/16 = fifteen and one-half\n\nAs x approaches 0 from either side, the value of the function approaches negative ∞.\n\nAs x gets very large, the graph approaches y = 4x from below as an asymptote. You can see that even by the time you get to x = 4, the graph gets pretty close to y = 4x.\n\nGraph those values. You don't need calculus.\n\nNice explanation by Wal. Openpsych has the value of f(-2) wrong.\n\nLet's take the first derivative of this and see where the curve turns.\n\nf'(x) = 4 + 16 / x^3\n0 = 4x^3 + 16\nx^3 = -4\nx = -cube root(4)\n\nThis is a local maximum. For x less than -cube root(4), the curve approaches x = 4y from below. For x greater than -cube root(4), the curve approaches negative  ∞.
762440
Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
944543
Okay . let see \n\nPart 1 \nVertical asymptotes correspond to the zeroes of the denominator of a rational function \nx^2+2x+1= 0 \n(x + 1)(x + 1)= 0\nx = -1 \nV.A = -1 \n\nPart 2 \nhorizontal asymptotes ; limit of f(x) as x → ∞\nlim (2x - 4) / (x^2+2x+1)\nx → ∞\nlim 2x / x^2   { 1 < 2 } \nx → ∞ \n= 0\nhorizontal asymptotes is " 0 "\n\nGoodLuck
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
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
862673
since there is no real answer to the square root of any negative number we write i\ni=the square root of -1\n\nif you did the square root of -25 you would get 5i which equals to 5 times the square root of -1
781968
I'm not sure if your question meant f(x)=2x(x-1)(x-4) or f(x)=x(x-1)(x-4)^2, so I will answer both.\n\n1) f(x)=2x(x-1)(x-4)\n(0, 1) U (4, ∞)\n\n2) f(x)=x(x-1)(x-4)^2\n(-∞, 0) U (1, 4) U (4, ∞)
875938
5\n\ntake the highest power in the polynomial, that exponent is your leading exponent, and is used as your "n"\n\nso 5 zeros.
1169084
Domain lists all possible x values. A normal polynomial function has a domain of all real numbers, because EVERY x value works.\n\nYour task is to find x values that DON'T work...and since you cannot divide by zero, then any x in the denominator that makes x^4+1 equal to zero will not be part of the domain.\n\nSo, what do you know about x^4+1?\n\nIt is an even power, so even if you put a negative number in for x, x^4+1 will be positive (since adding one will only make it more positive). If x=0, that is still positive, because of the plus one.\n\nx^4+1 is ALWAYS positive for any x...therefore it is NEVER equal to zero. Therefore any x value works, and your answer is all real numbers!!