Subject: Quandry


Math Is Book 6 (Ebos/Tuck) question 5(b) gives a surprising answer at the back of the book..

The question is:
Indicate the domain and range for this: y = sqr(x-9) (Square root of)

The answer for the domain is x>=9 which is not surprisingly, but the range is y>=0, yER ... which I could understand if you were not allowed to transform the equation into x as the subject..

But all example on the facing page DO TRANSFORM the equation at will with x and/or y as the subject..

Can you explain this...

Karel Marek

Hi Karel

Most of your difficulty seems to come from the lack of clarity of the text you are reading. What the text fails to tell you is that THERE IS NO GENERAL METHOD FOR FINDING THE DOMAIN AND RANGE OF A FUNCTION (except perhaps having a computer draw the graph, and your reading the domain and range from the computer screen!). We are often safe when we agree that a function will have as its domain any x-value where it is defined. We usually need to know what the range is only when we must find the inverse of a function. (Finding the inverse is what you refer to as "transform the equation into x as the subject". That is, if possible, taking the expression y = a function of x and solving for x to get x = a function of y.) Finding an inverse is usually such a difficult problem that we illustrate the process only with the simplest of examples. (In most cases it is easier to find the range than it is to find an inverse function, so I would not trust anybody who told me that in order to find the range of f(x) I should first find its inverse!)
So to answer your question, you must deal with each new type of function as you meet it.
THE SQUARE-ROOT FUNCTION (restricted to real numbers): Since you can take the square root only of a nonnegative number, the domain will include those x-values that make positive (or 0) the expression under the square-root symbol:

y = SQRT(x2 + 1) any x is OK
y = SQRT(x2 - 1) keep x bigger or equal 1.
y = SQRT(sin x) keep x between 0 and pi, or 2*pi and 3*pi, etc.
To find the range in these examples, you know that (by the definition of the symbol) the y-value must be positive, so you must be able to treat questions like
Can we make y zero? Can we make y arbitrarily large? Are there jumps in the y-values?
For the first example, x2 + 1 does not get smaller than 1 so that the smallest y-value is SQRT(1) = 1; since you can make x2 arbitrarily large, you can make x2 + 1 large, and so y can be arbitrarily large. Common sense tells us that y takes all values greater or equal 1.
In the next example, x2 - 1 can be 0, thus so can y, and the range goes through all nonnegative numbers.
In the final example, sin x runs between 0 and 1, and so does the square root.
THE QUOTIENT FUNCTION. The other major source of elementary examples is y = a quotient. The thing to avoid here is dividing by 0. To find the domain, it is important check whether the denominator can be zero. When the quotient is in "lowest terms", then any x-value that avoids division by 0 is allowed. Examples:

y = 1/(x2 + 1) any x is OK.
y = 1/(x2 - 1) any x except 1 and -1.
y = 1/sin x avoid any multiple of pi.
Finding the range for a quotient depends heavily on how complicated the numerator and denominator happen to be. If the denominator can go to 0 then the range will PROBABLY go off to plus or minus infinity (or both). Since x2 + 1 can't get below 1, y = 1/ (x2 + 1) can't go above 1, so it is forced to stay between 0 and 1 (including 1, not including 0). Since x2 - 1 can go from 0 to infinity on the positive side, y = 1/(x2 - 1) can take all positive values. By substituting values like .9 and .99, one easily sees that y goes off to negative infinity. But how does a person know that the range skips y-values between -1 and 0? I know it because I can draw the graph, but I don't know an easy method that will automatically show me that y cannot take values between -1 and 0. Obviously I can make such problems arbitrarily difficult by throwing extra terms into the numerator and denominator, and adding that to some other complicated expression. The range for y = 1/sin x (= csc x) are those values of y bigger than or equal to 1, and less than equal -1. If you need further convincing that the general problem is difficult, guess the range of y = (sin x)/x -- would you guess y stays between -1 and 1, even though there is a potential division by 0?
Finding the domain and range for functions made up from log x and the inverse trig functions is even harder -- probably impossible without seeing the graph of the function.
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