# Homework Help: Domain of the function

1. Jul 11, 2013

### nil1996

1. The problem statement, all variables and given/known data

f(x)= √[(2x+1)/x3-3x2+2x]

2. Relevant equations

3. The attempt at a solution

here
For f(x) to exist
x3-3x2+2x >0

and 2x+1≥0
by solving this i found that x $\in$ [-$\frac{1}{2}$,-∞]$\bigcup$[2,∞]
but the answer given is (-∞,-1/2]$\bigcup$(0,1)(2,∞)

so please anybody can explain why that is so

thanks

2. Jul 11, 2013

### Staff: Mentor

What about 2x+1 <= 0, x3-3x2+2x < 0?

There are brackets missing in the problem statement.

3. Jul 11, 2013

### HallsofIvy

I think you mean √[(2x+1)/(x3-3x2+2x)]

No, the numerator and denominator do NOT have to both be positive. What is true is that the fraction cannot be negative which means that, as long as the numerator is not 0, they must have the same sign. That is, either x3-3x2+2x >0 and 2x+1≥0 OR x3-3x2+2x <0 and 2x+1< 0.