How do you find the domain of $f(x)=sqrt(2x^2-5x+2)$ ?

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Answer 1 · 2017-04-10T09:24:16

$D=R$ without ${1/2, 2}$

$2x^2-5x+2>=0$

$b^2-4ac=25-4*2*2=9$
$sqrt(9)=3$

$x_1=(5-3)/4=1/2$
$x_2=(5+3)/4=2$

$D=R$ without ${1/2, 2}$

How do you find the domain of f(x)=sqrt(2x^2-5x+2)f(x)=sqrt(2x^2-5x+2)?