if f(x) = x/ (x +1) and g(x) = 1&#92;x, find f(g(x)) and state its domain?
917328
Gopal is correct, except (a) he meant x =  -1should be excluded from the domain, and (b) x = 0 should also be excluded from the domain (since g(0) is undefined).
754001
the domain and the range of h(x):\nthe domain : x is less than or equal to 2,the range : y is greater than or equal to zero.\n\nDomain and range for the inverse of h(x) :\nD: x is grater than or equal to zero.\nR: y is less than or equal to 2.\nNotice that the range of h(x) is the same as the domain of   the inverse  of h(x).And the domain of h(x)is the same as the range of the inverse of h(x).
575758
In the first problem, where you know g(x) = x + 4; and \nf(x) = 10x;  if you want to find g(f(x)), that means in the expression g(x), substitute 10x where ever you see x.  If you want to figure out f(g(x)), you would substitute (x + 4) for x in the expression 10x, or you would have 10(x + 4) = 10x + 40.\n\nI am going to assume for the second problems that the exponents are -5 and 3/4 respectively.\n\nf(g(x)) = 4(x^(3/4))^-5 = 4x^(-15/4)\n\ng(f(x)) = (4x^-5)^3/4 = 4^(3/4) x^(-15/4)
838016
This is basically just asking what is f(x)/g(x)\n\nso your answer would be \n\n(x+2) / (3x^2 - 5)
1166634
Say the point (1,2) is on f(x) and the point (2,5) is on g(x).\nNow since (1,2) is on f that means f(1) = 2.\nSimilarly, g(2) = 5 since (2,5) is on g.\nYou can now find (gof))(1)\n(gof)(1) = g[f(1)]  = g{2) = 5\n(1,5) is a point on (gof)(x)\nThe y-value of one function has to match the x-value of the other one.\nLet (4,7) be on f and (3,4) be on g.  You can find (fog)(3)\n(fog)(3) = f(g(3)) = f(4) = 7\n(3,7) is a point on (fog)(x)
753081
The domain of a function is all of the values of x that make the function "make sense."  In these equations, any x will work, so the domain is all x's.\n\nFor example, think about f(x) = sqrt(x).\nIt wouldn't make sense if x is -2!  \nThe domain of that function is all x >= 0.
951524
g(x)=f(2(x+4)\n=in f(x) whereverx is there sustitute 2(x+4)\nso the answer is 2x+8\n\n2.g(x)=f(-(x-2)\n=in f(x)=x^2 wherever there is x substitute\n-(x-2)\nso [-(x-2)]^2 will be your answer\nwhich is equal to\nx^2-4x+4\n\n3.g(x)=2f(-x)-5\nin f(x)=1/x,substituting\nf(-x)=-1/x\n2f(-x)=-2/x\nso -2/x-5 is your answer
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
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
71756
Here's a classic:\n\nUsing the following partial deriviatives:\n\nInt(f(x)*g(x))dx = f(x)*G(x) - Int(f'(x)*G(x))dx\n\nwhere G(x): the primitive function of g(x)\n     and f'(x): the derivative of f(x)\n\nNow watch the following 'proof':\n\n    Int(1/x^2 * 2x)dx  =  1/x^2 * x^2 - Int(-2/x^3 * x^2)dx\n           {just take f(x) = 1/x^2 and g(x) = 2x}\n\n    this yields:\n\n    Int(2/x)dx  =  1 - Int(-2/x)dx  = 1 + Int(2/x)dx    \n\n    substracting Int(2/x)dx on both sides yields:\n\n    0 = 1
865303
1) If h(x) = f(g(x)), then g(x) = 4+3x and f(x) = x^5\n2) Assuming your function is 3sqrt(x^2 -8), then\ng(x) = x^2 - 8 and f(x) = 3sqrt(x)
210475
f(g(x)) = 3/( (3/x)+4) -4)\nso that becomes\n3/ ((3/x)) because the + 4 and the - 4 cancel out\nand when you divide you multiply by recipricol so 3 divided by 3/x = 3 multiplied by x/3\n3 multiplied (x/3) = 3x/3 = x
974938
I believe this is how you do it:\n\n(f o g)(x) = f[g(x)] = 3(-x^2) - 4 = -3x^2 -4\n\nNow that this formula is found, sub in 2;\n\n(f o g)(2) = -3(2)^2 - 4 = -3(4) - 4 = -12 - 4 = -16
953356
Let f(x) = sin (1/x) and g(x)=-sin (1/x). Neither f nor g have limits at 0, but [x→0]lim f(x) + g(x) is obviously 0.
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
11252
The best way to find domain and range is to graph the function. If you graph the function that you have given as an example, you will notice that it is undefined at x=0. Since it is undefined at this point, the domain of the function is the real numbers minus 0. The range in general is slighly more difficult to find, but in this case is easy to find. This function goes to positive infinity when x approaches 0 from the left, and negative infinity when x approaches 0 from the right. The range of this function is all real numbers.
968502
Domain\n(-∞, ∞)\n\nRange\n[-3, ∞)