HELP !! convert    f(x) = x^2 + 1 (general form)  into standard form?
920068
The standard form of a quadratic:\n\nax^2 + bx +c\n\nIn your case a =1, and c=1. Since there is no x term, b=0
158205
Base on the Mathematics for STPM, Penerbit Fajar Bakti SDN.BHD. ,written by Ong Beng Sim...From the cphater 5,Coordinate Geometry, the actual standard form is ax/2+y/2=1.\n\nTo change the form:\n\nfind the axis-y & axis-x first,\n\nwhen y=0, x=2/a (axis-x)\nwhen x=0, y=2   (axis-y)\n\nBase on the standard form of linear funtion is x/a+y/b=1, where, a is axis-x and b is axis-y.\n\nso, substitute the axis-x & axis-y into the standard form of linear function...\n\nTherefore, the standard form of the linear funtion for y=-ax+2 is ax/2+y/2=1
34052
I do not know a preprogrammed utility to do this, but can point you in the direction that you might need to go.  See the reference below for the math and an on-line converter:
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
700913
a function is written in  the form f(x) or g(X)  or anything relating to that form\nan equation has and equal sign\n\ntherefore this one is a linear equation
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
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?
506639
The first derivitive of f(x) = 5/x^3 - 3/x^1/3 is...\n\nf'(x) = -15/x^4 + 1/x^(4/3)\n\nHope this helps. I used the product rule. Remember, you can write the term 5/x^3 and 5*x^-3. I find it simpler to rearrange the equation and use the product rule to get at the first derivitive.
757145
Lets see what values we can have for f(x). Suppose f(x) = c has some solution for x.\nThen (x^2 - 1) = c(x+2), so x^2 - cx - (2c+1) = 0. For that to have a solution, we need the discriminant, (b^2-4ac, though note thats a different c!) to be at least 0. So we need c^2+4(2c+1) >= 0.\nSo c^2 + 8c + 1 >= 0.\nNow, we can write that as (c+4)^2 - 15 >= 0, so (c+4)^2 >= 15.\nThus c+4 > sqrt(15) or c+4 < -sqrt(15).\nSo c > sqrt(15)-4 or c <-sqrt(15)-4.\n\nSo f(x) > sqrt(15)-4 or f(x) < -sqrt(15) - 4.\n\n(Note you could have done the last step solving the 'c' inequality with the quadratic formula; they're both the same, really.)\n\nNote that it is a much better idea to leave the inequality until as late as possible. Note that in the above idea of gp4rts, although it may be made to work, you *can't* do the following:\n(x^2-1)/(x+2) >= a\nTherefore\n(x^2-1) >= a(x+2).\nIf you multiply by a negative, you have to switch the inequality. Thus you would have to divide everything into two cases - if x+2>0, or x+2<0, and it gets very messy.
916581
To write any equation in "function form", just solve for y (get y by itself on one side of the =), and replace y with f(x).\n\nI'll let you do that.  :)
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
772631
Assuming this one is like your other questions, then\nAssuming h(x) = f(g(x)), then \ng(x) = x-1 and f(x) = 2x^5/3 + 5x^2/3
802168
The function is of the form d(u/v) which is given by\n(v du - u dv) / v^2.\nhere u=(1+x)^2, hence du=2(1+x)\nhere v=1-x, hence dv=-1.\nsubstitute these values in the formula and simplify to get final result.\n\nKSR
753779
When you have written it out twice you can then multiply out the brackets\ni.e. x^2(x^2-1/x^2)-(1/x^2)(x^2-1/x^2)\nthis will give you x^4-2+1/x^4\nRewrite using a negative indice to replace the fraction\nx^4-2+x^(-4) and then integrate as per norm using the rule\nthat the integral of x^n = x^(n+1)/(n+1).\nHope your day improves.
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)