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Mathematics
OpenStudy (anonymous):

Determine the domain or range of a function---I need help!

9 years ago
OpenStudy (shadowfiend):

Can you post which function you're trying to find the domain/range of?

9 years ago
OpenStudy (anonymous):

I tell my students, domain is all values of x that don't "cause problems" for the function. Problems include negatives inside square roots, dividing by zero, etc. A function like y = 2x + 1 has no problems, so the domain is all real numbers. A function like \[y=1/(x^2-1)\] can't have division by zero, so we must exclude 1 and -1; therefore, the domain is all real numbers except 1 and -1. A function like \[y=\sqrt{4-x^2}\] has a problem whenever 4 - x^2 is negative, so all those values of x must be excluded. Range is for y, and it's all the y values that are produced by the function. Suppose you have the function y = x^2, a parabola. Since all squares are positive, there is no way for the function to produce a negative value for y. Therefore the range is all non-negative real numbers. That help?

9 years ago
OpenStudy (anonymous):

Well my problem is: What is the domain of the function f(x)= 1/3 x + 3 when the range is {0, 3, 6} ?

9 years ago
OpenStudy (shadowfiend):

So in this case, you have the values of \(y\) (or in this case, \(f(x)\)), and you want to find the valid values of \(x\). To do that, you just solve the three equations with \(f(x)\) set to each of the three values they give you for the range, and the values of \(x\) for those equations will be the domain of the function.

9 years ago
OpenStudy (anonymous):

So I have to replace f or x with one of the three numbers {0,3,6} ?

9 years ago
OpenStudy (shadowfiend):

f.

9 years ago
OpenStudy (anonymous):

Okay thanks!

9 years ago
OpenStudy (shadowfiend):

No problem!

9 years ago
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