finding range of functions

Started by 113zami
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113zami

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how can you find the range of these functions:

1) y=x/1+x

2) y=x^2/1+x^2

3) y=x/sqrt(1+x^2)

4) the domain of the function y=sqrt(-x+9) is (-infinity,9] i understand the 9 part but why is it -infinity shouldn't is it be +infinity?

thanks
 
how can you find the range of these functions:

1) y=x/1+x

2) y=x^2/1+x^2

3) y=x/sqrt(1+x^2)

4) the domain of the function y=sqrt(-x+9) is (-infinity,9] i understand the 9 part but why is it -infinity shouldn't is it be +infinity?

thanks
You consider every possible value of x and the range is all the y values you can get out of it.

So for #1, x can be everything except -1 (then it would be undefined). For x --> negative inf, the function approaches 1. For x --> positive inf, the function also approaches 1. As x approaches -1 from the left, the function approaches positive inf. As x approaches -1 from the right, the function approaches negative inf. So I believe the range is all real numbers except +1.

For #2 it's x^2/1+x^2. That means we'll only have positive numbers. Right away y > or = 0. As x --> negative inf, the function approaches 1. As x --> positive inf, the function approaches 1. There are no undefined points. The denominator is always greater than the numerator so the function never hits 1 or above. The minimum is 0 at x = 0. So the range is [0, 1).

For #3 it's x/sqrt(1+x^2). Square roots can't be negative but that won't matter here. The lowest value of the square root is when x = 0 and you get y = 0 for that. As x --> negative inf we have a large negative number over the square root of a large negative number squared. At large values the +1 doesn't really do much to change it and so when you take the square root you get essentially -x/x which is -1. So it approaches -1. When you take x --> positive inf you essentially get a large number over the sqrt of a large number squared which again approaches 1. Note that it never reaches 1. The denominator is always greater than the numerator so we're working in between -1 and 1. There are no undefined points. Range is (-1, 1).

For #4 the domain of the function y=sqrt(-x+9) is (-infinity,9]. Why? Because the square root must be > or = 0 at all times. You figured out that 9 works because -9 + 9 = 0 and that's valid. The other end of it is -infinity because when negated you have +infinity + 9 which is certainly positive. Any number > 9 will give you a square root of a negative number and that doesn't have any real answer.

Hope this helps.
 
wow, how much do you charge for tutoring : )
did you do all that in your head, cuz i was only able to see it after working out the actual graphs with picking x and y values, all these fuctions were in the same question, so obviously i can't graph each one of them on the test
is there a way to find the range algebraically?? like you know for the domain, its just the x value that makes the denominator zero or make what's under the radicle negative
 
i hope someone can hlep me alittle more with the above question, i have searched all over the internet for a way to do it without looking at the actual graph, but there is none
where are you streetwolf?