How do you find the domain of $f(x)=sqrt((2x^2+5x-3) )$ ?

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Answer 1 · 2018-01-05T04:16:52

$D_f=(-oo,-3]uu[1/2,+oo)$

$f(x)=sqrt(2x^2+5x-3)$

$D_f={AAx$ $in$ $RR$ : $2x^2+5x-3>=0}$

Let's find the roots of $2x^2+5x-3=0$

$Δ=b^2-4ac=25-4*2*(-3)=25+24=49$

$x_(1,2)=(-b+-sqrtΔ)/(2a)$ $=$ $(-5+-sqrt49)/4$ $=$ $(-5+-7)/4$

So

For $x$ $in$ $(-oo,-3)$
for example $x=-5$ $->$ $2*(-5)^2+5(-5)-3=50-25-3=22>0$

For $x$ $in$ $(-3,1/2)$
for example $x=0$ $->$ $2*0^2+5*0-3=-3<0$

For $x$ $in$ $(1/2,+oo)$
for example $x=5$ $->$ $2*(5)^2+5*5-3=50+25-3=72>0$

Therefore $2x^2+5x-3>=0$ $<=>$ $x$ $in$ $(-oo,-3]uu[1/2,+oo)$

and $D_f=(-oo,-3]uu[1/2,+oo)$

How do you find the domain of f(x)=sqrt((2x^2+5x-3) )f(x)=sqrt((2x^2+5x-3) )?