f(x)=x^2 -3 Find Domain and Range show in Interval Notation?
968502
Domain\n(-∞, ∞)\n\nRange\n[-3, ∞)
1051867
The range is:\n(-1)<x<=(1)\nx=1 is admissible
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.
837800
did u mean square root of (2-x) or 1/(2-x)??\ni wil solve both the cases\n\nif it was sqrt(2-x), the solution is like this\nto find out the domain, we need to find all the values of x for which the function is defined. for a square root function, the only requisite is that the term inside the square root should be non-negative.\nso, the condition is (2-x)>=0, ie., x<=2\nso, the domain becomes (-infinity,2]\nnote that parenthsis, ie., '(' or ')' means that the value is not included in the interval, where as square brackets, ie., '[' or ']' means that the value is included.\n\nnow, if the problem is 1/(2-x), the solution is:\nfor this function to exist, the denominator should be non-zero, which means that (2-x) should not be equal to zero.\nso, x should not be equal to 2\nthe domain is given by  (-infinity,infinity) - {2}\nhere, {2} means the set containing the element 2\nthe domain can also be denoted as  R-{2}, where R is the set of all real numbers.
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
800627
x = 3/(x-2)\n\n=> x(x-2) = 3\n=> x^2 - 2x = 3\n=> x^2 - 2x - 3 = 0\n=> (x-3) (x+1) = 0\n\nEither x-3 = 0 => x=3\nOr x+1 = 0 => x=-1
902917
y=1/x^2-3 => y+3 = 1/x^2 => x= +/-sqrt(1/(y+3))
867199
domain {3.1,0}\nrange={-6,8,-9}
169470
The domain is the whole set R (all real numbers) since your function is defined for every x from R.\nSimply, you can put any real number to be x and you will have  real number as an answer.
867304
f'(x)=-3+10x-3x^2\nf''(x)=10-6x\n\nmax and min are where f'(x)=0\n3x^2-10x+3=0\n(x-3)(3x-1)=0\nx=3,1/3\n\nto see if they are max or min use f''(x)\nf''(3)=-8   means concave down= max at x=3\nf''(1/3)=8   means concave up= min at x=1/3\n\ninflection points are where f''(x)=0\n10-6x=0\nx=5/3\nf''(0)=10 means concave up on the interval (-inf,5/3)\nf''(5)=-20  means concave down on the interval (5/3,inf)
448269
The domain is all real numbers x, except for x=0.\n\nYou just have to remember that the exponential function and all polynomial functions are defined on the entire real line, that is there domain is all real numbers. Therefore, the only problem that occurs is with the operation of division by x. Since you cannot divide by 0, x can never be 0. All other values are fine though.\n\nIN RESPONSE to what urtimud said below: sure you can take e^0, but if you let x=0 you are taking e^(-1/0) which is undefined. Therefore, x=0 is not in the domain of this real valued function. Note that e is being raised to the ((x^2-1)/x) power, because your screen may have chopped off that last part.
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;
952602
This is what always helps me.\n\nA function is a machine.\n\nThe domain is everything you can put into the machine so that it doesn't break.\n\nThe range is everything you can get out of the machine.\n\nIn the case of your function, there's only one possible value of x that will "break" the function, in the sense that you'd get an infinite answer.  That value is x=0, because then you'd have 9/0 + 7, and 9/0 is infinity.\n\nSo the range is {x | x is not equal to 0}.\n\nAssuming that you meant 9/x + 7, rather than 9/(x+7).  If you meant the latter, x cannot be equal to -7.
98350
limit as x tends to 3 ;( x^2 -3)/2 =(3^2 -3)/2 =(9-3)/2 = 3
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
387682
This expression is defined for every real number, domain is the whole set R, because it wouldn't be defined only for x^2+3=0, but x^2+3 is always greater than zero.
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18