Domain of f(x) for √[(2x+1)/x3-3x2+2x]: Homework Solution and Explanation

[nil1996]

Homework Statement

f(x)= √[(2x+1)/x³-3x²+2x]

Homework Equations

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

 

[mfb]

What about 2x+1 <= 0, x³-3x²+2x < 0?

There are brackets missing in the problem statement.

 

[HallsofIvy]

Homework Statement

f(x)= √[(2x+1)/x³-3x²+2x]

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

Homework Equations

The Attempt at a Solution

here
For f(x) to exist
x³-3x²+2x >0

and 2x+1≥0

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.

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 sothanks