The domain of the function f(x) = √[(2x+1)/(x³-3x²+2x] requires that the expression inside the square root is non-negative. This leads to the conditions x³-3x²+2x > 0 and 2x+1 ≥ 0. The correct domain is (-∞, -1/2] ∪ (0, 1) ∪ (2, ∞), as the numerator and denominator must have the same sign, allowing for specific intervals where the function is defined. Misinterpretation of the conditions led to confusion regarding the domain boundaries.
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I think you mean √[(2x+1)/(x3-3x2+2x)]nil1996 said:Homework Statement
f(x)= √[(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.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 sothanks