f(x)=-sec(x) for 0<=x<=2pi determine domain and range?
953975
The domain is [0, pi/4), (pi/4, 3pi/4), (3pi/4, 2pi]\n\nThe range is [-1, -infinity), [1, infinity).\n\nJust in case, [ means "number next to it included" and ( means "number next to it not included." so [a, b) means from a (a included) to b (b not included.)
745315
domain=any real number
794327
A pretty trivial one would be f(x) = absolute value of x and g(x) = -x\n\nf(g(x)) = abs (-x) = x\n\nbut g(f(x)) = - (abs x) = -x
775550
The variable x can take any value so the domain is from - infinity to + infinity in real space (so any real number will do) so long as x-3>0 x>3. if not say good bye to the real domain since we end up with a negative number under the square root. So we have 3< x < infinity is the domain you are looking for\n\nExplanation: By definition the domain of a real function ( f(x) in our case) is the set of all real numbers variable x can take such that the f(x) remains in the range of real numbers.  \n\nReference http://www.analyzemath.com/DomainTest/answers_domain.html
1180422
The domain for a function is all the "x" values that work in the functions.\n\nAn example is "y = x"\nIn this case you can plug anything in for "x" and get a "y"\n\nAnother example is "y = 1/x"\nHere, x cannot be zero (since you cannot have 1/0)\nTherefore the domain is all values except x = 0\n\nIn your equation, (1/x^2)  - 3x  + 2, the domain is all real numbers except x = 0.  If x were 0, you'd have (1/0) - 0 + 0 and you cannot evaluate that.
1298
derivative of 5(sec(x)^2) with respect to sec(x) = 10(sec(x))\n\nderivative of 5(sec(x)^2) with respect to x = 10 * derivative of sec(x) with respect to x
764433
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
860983
sec pi\nsec(pi-0)\n-sec0\n=1/cos0\n=-1
790936
x≠０\ncuz if x=0, (x-7)/x^3 is meaningless
694986
domain is the value of x\nthat is {5,-3,4,-6.....}\nrange is the value of y\nthatis{2,4,2,3....}
840531
f(x) = -3x^2 - 6x - 5\n\ny-intercept (0, -5)\nno x-intercept\n\nvertex (-1, -2)\n\naxis x = -1\n\ndomain (-∞, ∞)\n\nrange (-∞, -2]
968502
Domain\n(-∞, ∞)\n\nRange\n[-3, ∞)
1052329
d, the domain is the set of all real numbers except for those values where the function is undefined, such as when evaluating the function would require you to divide by zero.\n\nFor the purposes of this question, only the (x^2 - 49) component is relevant. When x is seven or negative seven, this term will be zero, making the entire denominator zero, and thereby making the function undefined.
126637
http://myweb2.search.yahoo.com/search?p=%22difference+between+a+function+and+an+equation%22&ei=UTF-8&fr=slv1-msgr&x=wrt
768749
A'ighty, first write sec(x) = 1/cos(x).  Now take the derivative using the Quotient Rule:\n\nd/dx (1/cos(x)) = (cos(x)*d/dx(1) - 1*d/dx(cos(x))) / cos^2(x)\n\n                        = (0 - (-sin(x))) / cos^2(x)\n\n                        = sin(x) / cos^2(x)\n\n                        = (1/cos(x)) * (sin(x) / cos(x))\n\n                        = sec(x)*tan(x).\n\nAnd there you go.  ^_^
940347
In this case, the domain is all real numbers, because you can use any number for x. And the range is all real numbers, because you can find an x that will produce any real number y. So you write the domain and range as follows:\n\nD = {x€R} (€ is the nearest I can find to the symbol for "is an element of")\n\nR = {y€R}\n\nor you can say \n\n-∞<x<∞\n-∞<y<∞\n\nHope that helps.
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!