What is the range of X/(X^2-9)?
826809
It is probably best to determine the range after you have the graph.  You will need to look at the local maximums, local minimums, and the horizontal asymptotes to determine the range.  (In other words, it is not a fast problem).
651993
Here's the best I can do to help with the first one.\n   __x_+9___\nx-2)X^2+7x-2\n    x^2-2x\n    ------\n        9x-2\n        9x-18\n        -----\n           16\nFollow the example for the rest
535346
I'm not sure if this is the whole question, but I will interpret it as if it ended with a ) in place of the [exp...\n\n9(x^3)N - 12(x^4)(N^2) - 9(x^2)N
914435
x > 5\n\n- - - - - -s-
500014
a. (x+9)(x-9)\nb.(t-2)(t+2)\nc. (b-12)(b+12)
780025
x ≤ 5\n\n-- - - - - - -s-
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
601402
depends on what you mean\nif (9x)^2 - 81 = 0 then\n(9x)^2 = 81\n81x^2 = 81\nx^2 = 1\nx = 1 or -1\n\nif 9x^2 -81 = 0 then\n9x^2 = 81 \nx^2 = 9\nx = 3 or -3\n\ntosmith
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.
1082686
Well this is a rational expression, so you have to comsider that it is undefined if the denominator equals 0.  So set the denominator equal to zero, solve for x, and we know what numbers can't be used for x.\nx^2-81=0  is a difference of two squares\n(x+9)(x-9)=0\nx+9=0 or x-9=0\nx=-9 or x=9\nTherefore possible solutions for x are all real numbers except -9,9.
239483
zero  you said squared, not to the zero power.  0^2=0  try it on your calculator.
790936
x≠０\ncuz if x=0, (x-7)/x^3 is meaningless
415757
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
203508
Two.\n\nYou can rewrite it as x^2 = 15 or x = sqrt(15)\n\nx solves to (rounded) 3.87 or -3.87
618060
this would multiply out to be 4x^2-9\n\nif you let the expression be equal to 0, then x would be positive or negative 3/2 because either (2x+3) = 0 or (2x-3) = 0, and just solve for x