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| Math Central | Quandaries & Queries |
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Question from Camille, a student: Hi There, I am taking a calculus class by correspondence and part of the introduction is learning how to find domain and range of functions. I understand how to find the domain. I also understand that if you find the inverse of the equation and solve for x to find the domain it will give you the range because the domain of the inverse is equivalent to the range of the original equation. However when trying to apply this method to complicated questions (ie: square root in the denominator) I get confused and can't solve for x. Here is a sample question: f(x) = 3 / sqrt(x-2) For this question I know that the domain is (2, infinity) because the division by zero is not allowed and because of the square root sigh it is only defined when x-2 > 0. How would I find the range for this question? Find inverse: y = 3 / sqrt (x-2) x = 3 / sqrt (y - 2) [sqrt (y - 2)] x = 3 This is where I get confused because how do I multiply out the variables on the left side of the equation? Thank you |
Hi Camille,
There is no general procedure for finding the domain or range of a function. You should read what Chris wrote in response to a similar question. The example you give shows one kind of trap you can fall into.
Find inverse: y = 3 / sqrt (x-2)
You interchanged x and y to get x = 3 / sqrt (y - 2)
You can solve this for y by squaring both sides
x2 = 9/(y - 2)
y - 2 = 9/x2
y = 9/x2 + 2
This function has a domain of all x except x = 0 so is that the range of the original function? Look at the original function, y = 3 / sqrt (x-2). You can see that y can't be negative. Thus for example -1 is in the domain of y = 9/x2 + 2 but it's not in the range of y = 3 / sqrt (x-2). What did I do wrong?
The trap I fell into is in squaring x = 3 / sqrt (y - 2). In this expression x must be positive and when I squared both sides to get x2 = 9/(y - 2) I have an expression where x can be negative.
Hence in reply to your question of how you solve x = 3 / sqrt (y - 2) for y, you square both sides but when you square you need to keep track of any restrictions that are contained in the original form. Hence I have
x = 3 / sqrt (y - 2); x > 0, y > 2
x2 = 9/(y - 2); x > 0, y > 2
y - 2 = 9/x2 ; x > 0, y > 2
y = 9/x2 + 2; x > 0, y > 2
The final form of the inverse has the property that x can't be zero but that is already contained in the restriction that x > 0 so the domain of the inverse is (0, ∞) and hence the range of the function y = 3 / sqrt (x-2) is (0, ∞).
I hope this helps,
Harley
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