For which of the following functions f is f(x) = f(1-x) for all x?
485166
The correct answer is d.\n\nf(1-x) will give a negative number for all positive 'x' values.\n\nbut squaring these negative number will always give a positive number i.e (-4)^2 = +16.\n\nTherefore in the equation f(x) = x^2(1-x)^2;\n\nYou can see that for all positive or negative numbers you will have a positive value. Again,\n\nLet x=5; Then 1-x = -4;\n\nFor the equation d.. 1-(-4) = 5. i.e the original number.\n\nHence, f(5) = 5^2 (-4)^2 = 25*16;\nAlso, f(-4) = (-4)^2 (5)^2 = 25*16;\n\nThus, for all values of x... Equation d is TRUE.
768435
(fog)(x) and (gof)(x) are called composite functions of x.\nnow, in the first case, f(x)=2x-3, g(x)=x+7\nso, (fog)(x)=f(g(x)), ie., we have to substitute g(x) in place of x in  f(x).\nso, f(g(x))=2(g(x))-3\n                =2(x+7)-3\n                =2x-11\nsimilarly, gof(x)=g(f(x))\n                         =(2x-3)+7\n                         =2x+4\nin the second problem also, we hav to follow the same approach\nso, fog(x)=f(g(x))\n               =(3x+1)^2+(3x+1)-2\n               =9x^2+1+6x+3x-1\n               =9x^2+9x\nsimilarly, gof(x)=g(f(x))\n                        =3(x^2+x-2)+1\n                        =3x^2+3x-5
904826
f(g(x)) = g(x) + 1 = (x^2 - 1) + 1 = x^2  - 1 + 1 = x^2\n\ng(f(x)) = (f(x))^2 - 1 = ( x + 1 )^2 - 1 = (x^2 + 2x + 1) - 1 \n                                                         = x^2 + 2x + 1 - 1 \n                                                          = x^2 + 2x
931057
f(x) = sin½x + a \nSo x = sin(½f^(-1)( x)) + a\nx - a = sin(½f^(-1)( x))\nso ½f^(-1)( x) = arcsin(x - a)\nSo f^(-1)( x) = 2 arcsin(x - a) \nDomain: a -1 ≤ x ≤ a + 1\nRange: - π ≤ f^(-1)( x) ≤ π
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.
1102849
gurl am not helping out with ur homework next time.\n f(x) =/x/ +3x +1\n       =x+3x+1\n       = 4x+1\n\n  f(x)=3x+/2x-5/\n       = 3x+2x-5\n       =5x-5\n\n  g(x)= /x/+/x-1/\n        = x+x-1\n        =2x-1\n\n  g(x)=3/x-2/ - /x+1/\n        =3(x-2) - x+1\n        =3x-6-x+1\n        =2x-5
71723
You just plugin one function into another function where there's an X.\n\ng(f) means use function G and plugin function F to replace the X\n\n1.  g(f)=3-2x f(g)=2(3-x)=6-2x\nI just replaced 2x with x for g(f) and 3-x for x for f(g)\n\nUse this method to figure out the rest.\n\nNotice that it says "x is not equal to whatever".\nThis just states that it will be undefined if it is that number and for now, should be ignored as all problems will have real answers.
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
437573
Functions that maps x to x\nWhere x is an element of the function
966844
f(0) = -i\nf(1) = 1² - i = 1-i\nf(2) = 2² - i = 4-i\n\nI take it you are dealing with complex numbers?  i is the square root of -1.\n\nNRT
150385
f(x)=Int(f'(y), 0, x)-f(0) => yes, they're opposite operation, but if x has a given value, e.g. x=2, 3...no, they're not, b/c in that case, the RHS of the previous egality is a real number, not a function
770442
1)  No, in fact constant functions are never one-to-one.  The requirement for a one-to-one or "injective" fuction is that f(a) = f(b) if and only if a = b.  With a constant function, f(a) = f(b) for all a and b, even when a != b.  (That's != for unequal.)\n\n2)  No, because f(a) = f(-a) for all real numbers, and a != -a unless a = 0.  Note that p(-2) = p(2) = 5, but we couldn't have p(-2) = p(2) for an injective function because -2 != 2.
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
207605
bool function(int a, int b, int c)\n{\nreturn a || b || c;\n}
669001
e^f(x) = x is the best answer i can give you.
388704
it cannot be f(x) = a for all x, unless a=0; otherwise you would be pretending that a = a+a with a non-0 !! This doesn't happen in R.
760867
f(x) = (x+3)(x^2 + 16)