What is the domain of #f(x) = (3x-1) / (x^2-x) #?

1 Answer
Stefan V.
Sep 7, 2015

Domain: #(-oo, 0) uu (0, 1) uu (1, + oo)#

Explanation:

Right from the start, you know that the domain of the function must not include any value of #x# that will make the denominator of the fraction equal to zero.

So basically, any value of #x# for which

#x^2 - x != 0#

wil be excluded from the function's domain.

You can rewrite this equation like this

#x^2 - x = 0#

#x * (x-1) = 0 implies {(x=0), (x=1) :}#

This means that the domain of the function will be #RR-{0, 1}#, or, in interval notation, #(-oo, 0) uu (0, 1) uu (1, + oo)#.

graph{(3x - 1)/(x^2 - x) [-46.22, 46.22, -23.1, 23.15]}

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