RangeDate: 04/02/2002 at 14:08:23 From: Teresa Otworth Subject: Range If f(x) = x-3/x+3, I know the domain to be all Reals except for -3. The range, however, is all Reals except for 1. I don't know how they get the 1 algebraically. Thank you for your time. Teresa Otworth Date: 04/02/2002 at 17:08:26 From: Doctor Peterson Subject: Re: Range Hi, Teresa. The usual way to do this is to try to invert the function; then the domain of the inverse is the range of the function. That is, we solve for x: y = (x-3)/(x+3) y(x+3) = x-3 xy + 3y = x - 3 xy - x = -3 - 3y x(y - 1) = -(3 + 3y) x = (3 + 3y)/(1 - y) Now we can see immediately that y can be anything but 1. Since for any y except 1, there is an x, we know that the original function can take any value except 1. Having done that, we can look back at the original function and see why it can't be 1. If we had (x-3)/(x+3) = 1 then x-3 = x+3, which is impossible. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/04/2002 at 13:59:29 From: Teresa Otworth Subject: Range In f(x) = 2/sqrt(x-4), how do you find the range? Date: 04/04/2002 at 15:53:36 From: Doctor Peterson Subject: Re: Range Hi, Teresa. Did you try using the method I showed you last time? We can help a lot better if you show us your work so we can see why you are having trouble - just as a medical doctor wants to see where it hurts, a math doctor wants to see what doesn't work. Solve y = 2/sqrt(x-4) for x, and see what values of y do not allow you to find an x. Those values are excluded from the range. In this case there is an extra issue: the square root is always taken as the POSITIVE root, which limits y further. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/04/2002 at 19:28:36 From: Teresa Otworth Subject: Range Thanks for being patient with me. I worked f(x)= 2/sqrt(x-4) like this: y(sqrt(x-4)= 2. Then, I moved the y: sqrt(x-4)= 2/y. Then I squared both sides and obtained x-4 = 4/y^2 This is where my question arises. If I then move the four on the left side of the equation and then solve for one y, I get an imaginary number. Was that what you were referring to that would always be positive? That would give a 4 + 4, so I am lost at that point. Date: 04/04/2002 at 22:18:50 From: Doctor Peterson Subject: Re: Range Hi, Teresa. You've done well so far. Remember you are solving for x, and you get x = 4 + 4/y^2 You don't want to solve for y now; the purpose of solving for x was to see for which values of y you can find an x. Look at this equation, and you see that you can put any value of y except for 0 into it; the x you get out is just the x you could put into the original equation to get that value of y. So f(x) can take any value but 0. But ... Let's try an example or two. If y = 1, we get x = 4 + 4/1 = 8, and f(8) = 2/sqrt(8-4) = 2/2 = 1 so we did in fact show that 1 is in the range. Now let y = -1. We find that x = 4 + 4/(-1)^2 = 4 + 4/1 = 8 But that's the same x we got for y = 1, and we know that f(8) is not -1. So what happened? When you square y, both 1 and -1 give the same result. Another way to say that is that 1 and -1 are both square roots of 1. But in the original equation, we had the square root _function_, which is defined as the POSITIVE square root only. So f(x) = 2/sqrt(x-4) always gives the positive root, and so f(x) has to be always positive. That's why we don't get -1 out of it when we put 8 in, even though both 2 and -2 are square roots of 4. So the answer is that the range of this function is all positive real numbers: zero is excluded because we couldn't divide by y^2 in our inverse, and negative numbers are excluded by the square root function. I've gone through the whole thing, because this last part is tricky. Let me know if you don't follow any part of it. By the way, a good way to get a better feel for these problems is to try graphing them after you've worked out the domain and range by other means. Do that in this case and you will see what's going on much more clearly. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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