∫ [f(x) / (1 + x)^2]dx, where f(x) is a polynomial of degree two in x, such that f(0)=1,f(1)=3,f(-1)=5.?
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
565133
We start by factoring the polynomial.  A little trial and error (testing the factors of 2, which is the constant term) reveals that 2 is a root.  So, we know that (x-2) is a factor.\n\nDividing the polynomial by (x-2) using synthetic division, we find the other factor is 6x^2+x-1.  This can be factored as (3x-1) (2x+1).\n\nSo, the equation, in factored form, is\n(x-2)(3x-1)(2x+1)=0\n\nSo, the roots are x=2, x=1/3, and x=-1/2.
465550
Do you mean "zero of a polynomial?" If so, then read on...\n\nA zero of a polynomial is the value or values of the variable (usually x) that cause the polynomial to equal zero.\n\nExample 1: The zero of the monomial 3*x is 0 because 3*0 is zero.\n\nExample 2: The binomial x + 2 has it's zero at x = -2 because if you substitute -2 in place of x, then you get \n---> -2 + 2 = 0.\nTherefore, -2 is the zero of the given binomial.\n\nExample 3: Find the zeros of the trinomial x^2 - x - 6. That is, solve x^2 - x - 6 = 0. If you factor the trinomial, then you get \n---> (x + 2)(x - 3) = 0. \nIn this problem there are two possibilities. Either \n---> x + 2 = 0  or  x - 3 = 0. \nSolving each gives x = -2 or x = 3. Therefore, the zeros of the given trinomial are -2 and 3. \n\nIf you substitute -2 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (-2)^2 - (-2) - 6 \n= 4 + 2 - 6\n= 6 - 6\n= 0 \n\nIf you substitute 3 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (3)^2 - (3) - 6 \n= 9 - 3 - 6\n= 6 - 6\n= 0 \n\nThere are no other values of x that will cause our given trinomial to equal zero.\n\nDefinitions\nA monomial is a polynomial with one term.\nA binomial is a polynomial with two terms.\nA trinomial is a polynomial with three terms.\nA polynomial with more than three terms is just called a polynomial.\n\nNote: If you graph a polynomial, then the graph will touch or cross the x-axis at the zeros of the polynomial. Zeros of a polynomial are also called solutions, roots, and, in some cases, x-intercepts.
824215
Did you mean -(3/4) * (x^3+2x^2-3x+1) ?\n\nLet f(x) =  -(3/4) * (x^3+2x^2-3x+1). Equivalently, f(x) = -(3/4)x^3 - (3/2)x^2 + (9/4)x - 3/4. 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\nSince f is a polynomial function (of degree 3) on x, it follows that this is continuous on the set of real numbers. Thus, by applying the limit on both sides (left and right) of x, we will obtain the same value for both sides. \n\nThat is, the left-hand Limit will be as follows: as x approaches to the left of c, we make f(x) approaches f(c) = -(3/4)c^3 - (3/2)c^2 + (9/4)c - 3/4. Similarly, the right-hand limit will be: as x approaches to the right c, we also get f(x) approaches to f(c). \n\nTherefore, both the left and right-hand limit yields to f(c); so as x approaches c in any direction, we always obtain f(c). Hence, both left and right-hand behavior of f are the same for any choice of a real valued x.
986272
f(x) = 3(2/x + 1)\ng(x) = x^2\n\n\nHope this helps.
317595
If f is an antiderivative of (tanx)^2/(x^2+1) such that f(1)=0.5, then f(0)=__?\n\nnumerical integral of (tan x)^2/(x^2 + 1) from 0 to 1 = 0.345\nf(x) = numerical integral of (tan x)^2/(x^2 + 1) from 0 to x + 0.155\nso f(0) = 0.155
195265
lim h->0 (f(x+h)-f(x))/h\n\n[(x+h)^2-3(x+h)+1-(x^2-3x+1)]/h\n=(x^2+2xh+h^2-3x-3h+1\n    -x^2+3x-1)/h\n=(2xh+h^2-3h)/h\n=(2x+h-3)\n\nThere should be a limit as h->0 in front of each line above, so let h=0 in the final line to give you 2x-3.\n\nThis is called a derivative by the way, and if you haven't learned them yet there are easy rules to follow for computing the derivative of polynomials like this one.
735529
f(x)=-2x^4+6x+7\nf(-x)=-2(-x)^4+6(-x)+7\n=-2x^4-6x+7\nnether odd nor even
1048436
As a^2 is contained in F(a), clearly F(a^2) is contained in F(a). So we need to show the reverse inclusion.\n\nWe have the chain of fields F < F(a^2) <= F(a).\n\n(where I use <= and < for subset and proper subset containment) \n\nand hence [F(a) : F] = [F(a) : F(a^2)][F(a^2) : F]. Set n = [F(a) : F] and m = [F(a^2) : F]. The preceeding equation implies that m divides n. It will suffice to that m = n. Let f(X) = X^m + c_{m-1}X^{m-1} + ... + c_1X + c_0 denote the minimal polynomial of a^2 over F. Then f(X^2) = X^{2m} + c_{m-1}X^{2(m-1)} + ... + c_1X^2 + c_0 has a as a root. This implies that the minimal polynomial of a has degree less than or equal to 2m. So we get m divides n and n <= 2m. The only ways for this to occur are when m=n or when 2m = n. But n is odd, so n does not equal 2m. Hence m = n. This completes the proof.
173410
two ways.\n1) this equation is equal to the product:\n(2x-1)*(-2x+1)=0\nSo, x=1/2\n2) For a polynomial of degree 2, in the form of \n((a*x^2)+(b*x)+c).the discriminant is [(b^2)-(4*a*c)].Here, a=-4 b=4 c=-1.\ndiscriminant= 16-(4*(-4)*(-1))=0\nthe formulas for x are =[-b+((dis)^(1/2))]/(2*a) and\n[-b+((dis)^(1/2))]/(2*a).\n\nSo, x=[-(-4)+0]/(2*4)=1/2 and =[-(-4)-0]/(2*4)=1/2
715887
f(x) = x^5 +2x^3 -x^2 -2x -6\nf(x) has one change of sign so it has one positive root.\n\nf(-x) = -x^5 -2x^3 -x^2 +2x -6\n\nso it has got 0 or 2 -ve roots.\n\nThis is as per Descate's rule
54176
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
837305
f(x) = [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2) ]\n= 3(x - 2)/(x - 3) (x ≠2, -2)\n= [3(x - 3) + 3]/(x - 3)\n= 3 + 3/(x - 3)\n\n1. Lim x→2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→2 [3 + 3/(x - 3)]  \n= 0 \nand \nLim x→-2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→-2 [3 + 3/(x - 3)] \n= 13/5\nSo it has 'holes' at (2, 0) and (-2, 13/5)\n\n2. It is undefined when x = 3, 2 and -2\nSo domain is all x except x = 3, 2, -2\nRange all y except y = 0 and 13/5\n\n3. As x increases and decreases without bound \n3/(x-3) approaches 0\nso 3 + 3/(x-3) approaches 3\n\nie lim x → ±∞ [3 + 3/(x-3)] = 3\n\n4. When x < 3\nx - 3 < 0 \nSo 3 + 3/(x-3) < 3\n\nWhen x > 3\nx - 3 > 0 \nSo 3 + 3/(x-3) > 3\n\nSo for x > 3 the curve is above y = 3 and for x< 3 the curve is below y = 3\n\n5. When y = f(x) = 0\n3(x - 2)/(x - 3) = 0\nie x = 2\nSo f(x) cuts x-axis at x = 2\n\n6. When x = 0\nf(x) = 3*(-2)/(-3) = 2\nSo y-intercept is y = 2\n\n7. It is rotationally symmetrical about the point (3, 3) (order 2) except for the 'holes'.\n\nThat should get you a good start I think
863763
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
216211
when x=0 \ny=f(x)=\n(-1/15)(-5)(3)^2(-1)^3=\n(-1/15)(-5)(9)(-1)=\n(5/15) *(-9)=\n(1/3)*(-9)=\n-9/3=-3\n\n(0,-3) y-intercept\n\n\nWhen y=0\n\n-1/15(x-5)(x+3)^2(x-1)^3=0 multiply by -15\n(x-5)(x+3)^2(x-1)^3=0\n\nso x-5=0 and x=5  or\n   x+3=0 and x=-3 or\n   x-1=0 and x=1\n\nso (5,0) (-3,0) and (1,0) are x intercepts.
38209
actually polynomial functions make me happy...they are actually pretty easy.  You evaluate each term (thats the stuff separated by the addition (+) or subtraction (-) sign) individually.  So if you had sin(x) -x^3 + 3x\n\nthen you have the following\n\ncos(x) -3x^2 +3\n\nhope this helps!!