My Math Forum i am so stuck on these math problems

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 July 21st, 2008, 09:24 AM #1 Newbie   Joined: Jul 2008 Posts: 2 Thanks: 0 i am so stuck on these math problems [color=#FF0000]I dont expect for you to answer all of them but anything you do answer would be really appreciated, these are my study guided problems for a math final and i am basically stuck on this sheet theres 30 sheets and this is just one of them so you can imagine wat i have to do. THANKS IN ADVANCE![/color] 2. The domain of f where f(x) = 1/square root of 4-x^2 is _______ 6. f: A?B is a function from A to B and f(A) = B then f is called _______ 7. f = { (a,1), (b,2), (c,3) } then f is______ 11. If f = {(1, a) (2, b), (3, c) (4, d)} then its range is _______ 13. f is a function defined by f(x) = 3x^3-7x^2+6x+1/2x^3-3x+9 then f is called _______ 15. A function f(x) = [x] then f(-3/4) = _______ 17. f is a function defined by f(x) = -x/5 and g is a function defined by g(x) = 7-3x then 4g - 5f = _______ 20. The domain of f = 1/4x+1 is _______
 July 21st, 2008, 12:16 PM #2 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: i am so stuck on these math problems I'll start with the first and last problems about domain. Realistically speaking, you cannot divide by zero, nor can you take the square root of a negative number. Thus 1 / (4 - x^2)^.5 has restrictions. When x = 2, for example, we would be dividing by zero. When x is greater than 2, we would be taking the root of a negative number. Likewise, when x = -2 or when x is less than -2 we have problems. Thus the domain of the function, which is to say the input (x) values that can be inserted into the function, is from -2 to 2 (not including either). For the last one, there are no square roots. But we can have a situation where we divide by zero. In order to avoid that, 4x + 1 cannot be equal to 0...

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