how do you determine the domain (f of g)(x)?
400472
well for f of g, you know the answer is \n       5/ (4(square root of x) -2)    right? \nTo find the domain, you just need to find all the x values the function exists at-- (the word domain means x-values.)\nIt should exist everywhere except where\n    4(square root of x)-2 \nequals zero, because then the denominator would equal zero, and that is undefined.\n\nif you need more help just ask!
323916
sec is 1/cos. So the domain is the set of all real numbers except when cos(2x - pi) is 0. To graph it, first plot the cosine graph corresponding to -2cos(2x - pi). Then, "flip" the wave along y=2 and y=-2 for when the graph is positive and negative. Basically, what you get are sets of parabolas going off to infinity alternating between being above and below the x axes. Then, erase your cosine graph and shift these "parabolas" up by 3, and you're good.\nSince -2cos can take any value between -2 and 2, -2sec can take any value except those between -2 and 2. Shift up by 3, and you get a range of all real numbers less than or equal to 1 and greater than or equal to 4.
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
493069
First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
179824
Its preety easy. The answer is (3X^2)(3^X). Just take the derivative of X^3. Don't do any thing with 3^X cuz X is an exponent and when a X is an exponent you just leave it like that.
756998
f(g(x))= 4g(x) -5\n= 4(x/3) -5\n= 4x/3 -5\ns
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.
957165
The domain is the set of all values the variables can have so that the value of the expression is defined.  A rational expression has a definite value as long as the denominator is not zero, since division by zero is undefined.  So any set of values for the variables is in the domain just as long as that set doesn't make the denominator of the expression equal zero.  \n\nExample :  (x^2 + 5xy + 4) / (x - 3)(y + 7) is a rational expression.  You can calculate its value for any values of x and y you plug into it just as long as the denominator (x - 3)(y + 7) doesn't equal zero.   But (x - 3)(y + 7) = 0 if x = 3 or y = -7.  So any pair (x,y) is in the domain of the original expression unless x = 3 or y = -7.  \n(1,2), (5,10), (-100,100), (4/3, -1/2) are all in the domain.  \n(2,-7), (3,41), (-8,-7), (3,-7) are not in the domain.
785253
So are you looking for a function with a singularity on (0,1)?  That would address the question I think.\nmaybe (x-(1/2))^(1/2)\n\nTHE FIRST ANSWER IS WRONG, I THINK, BECAUSE YOUR DOMAIN IS OPEN AND DOESN'T INCLUDE 0.
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
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
851057
There is no work involved. You only need to understand what domain and range mean. Domain is what can you plug in for x, range is what numbers you can get out of the function.\n\na.\nDomain: all real numbers except -9, since f(-9) will have 0 in the denominator.\nRange: all real numbers except 0, since the numerator can never be 0\n\nb.\nDomain: all real numbers\nRange: y>=-4 (y is greater than or equal to -4). You have to find the turning point using x=-b/2a=-2. f(-2)=-4\n\nc.\nDomain: all real numbers except x=-1,3, since f(-1),f(3) have 0 in the denominator.\nRange: all real numbers.
867199
domain {3.1,0}\nrange={-6,8,-9}
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).
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.
730130
Domain = x... should be all real numbers\n\nRange = y... 2 and up... x can be a negative number, but will always end up being postive when added to two, therefore the lowest number is 2
457089
G=uys\nF=***\nE=verthing