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How do I find range and domain of a function?

I keep forgetting because I haven't learned it by heart yet. For example, how do I find the domain and range of this: y = (3x^2)/(x^2+1) Please tell me how to find domain and range, thanks.

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1. The domain is all reals, because x^2 + 1 > 0 and the only # that is not in the domain is an imaginary number. The range is also all reals. Correction: I totally forgot about the x^2 on the top and bottom. The range is therefore all positive integers from 0 to 3. I found an example: 3(10)^2 / ((10)^2 + 1) ~ 2.97
2. domain all R, x can be any number, since you don't have any restrictions in denominator range = [0...3] since you have square there won't be any negative y values for the same reason (square on the top and x^2+1) range is from 0 to 3 pick any number for x you'll see the function will never reach 3
3. range is the values of x you can put into the function. Since there aren't any square root or trig functions, the only values you can't put in would make denominator equal to zero. When you set the denominator equal to zero, you find there are no real numbers that will make x^2+1 = 0, so the demain is all real numbers For the domain as you go to plus and minus infinity and zero what happens to the output of the function? Well with the square both plus and minus infinity will do the same thing. The one in the denominator will become insignificant leaving you with 3x^2 over x^2 = 3 The lowest value happens at zero, which is 1. 1 to 3 is the domain.
4. How to find domain: You can do this by graphing. Domain is the "range" of x values in the graph. Like does it g from 0 to 10 on the x axis? Then your domain would be written as D= x| x (greater than or equal to symbol) 0 <x (less than or equal to symbol) 10, XER The range is the range of y values on the y axis. It is written as R=y|y(greater than or equal to symbol)write your lowest y value here<y(less than or equal to symbol) your highest y value, XER
5. Domain is probably the easier of the two. Domain will always be (-inf, inf) unless you're dealing with certain functions: 1. log_b ( X ); X > 0. 2. X ^ [1/a] where a is even; X >= 0. 3. 1 / X; X != 0 ( != means "is not equal to") These are the three basic functions with a domain restriction. As you're introduced to more functions, learn their basic domains as well. So, we don't have 1 or 2, but we do have 3. So, to solve for the domain, take whatever is being used as X, in this case "x^2 + 1", and then plug that in for X in the appropriate domain sentence x^2 + 1 != 0. Then you have to solve that inequality for whatever variable appears, in this case x. If there is no solution, then the domain is (-inf, inf). For range, it's a little more complicated. You should also learn the ranges of whatever basic functions you encounter. Your basic strategy is to find numbers that if you plugged them in for y, you couldn't solve for x. For a rational function like this, you have to remember that they have horizontal asymptotes. So find this asymptote. Set y equal to this asymptote and see if you can solve for x. If not, it's not in the range. Then in this case, note that the top and bottom of the fraction can both never be negative (but the top can be 0). So we know that (-inf, 0) is not in the domain, but [0, HA) (whatever you find that horizontal asymptote to be) is. Next, try to plug in a number greater than the HA for y, and see if you can solve for x. If you can, then (HA, inf) is in the domain, but otherwise not. Good luck!