What is the domain of f(x)?

Question

$f (x) = \frac{1}{(2 x - 1) + \sqrt{x^{2} - 3}}$ $f(x)= 1/((2x-1)+sqrt(x^2-3)$

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Answer 1 · 2017-12-31T18:16:06

$D (f) = (- \infty, - 3] ⋃ [3, \infty)$ $D(f)=(-oo,-3]uuu[3,oo)$

$I_{1} : (2 x - 1) + \sqrt{x^{2} - 3} \neq 0$ $I_1:(2x−1)+sqrt(x^2−3)!=0$
$I_{2} : x^{2} - 3 \geq 0$ $I_2:x^2-3>=0$

$D (f) = I_{1} ⋂ I_{2}$ $D(f)=I_1nnnI_2$

$2 x - 1 + \sqrt{x^{2} - 3} \neq 0$ $2x−1+sqrt(x^2−3)!=0$

$\sqrt{x^{2} - 3} \neq 1 - 2 x$ $sqrt(x^2−3)!=1-2x$

$x^{2} - 3 \neq {(1 - 2 x)}^{2}$ $x^2−3!=(1-2x)^2$

$x^{2} - 3 \neq 1 - 4 x + 4 x^{2}$ $x^2−3!=1-4x+4x^2$

$0 \neq 4 - 4 x + 3 x^{2}$ $0!=4-4x+3x^2$

$3 x^{2} - 4 x + 4 \neq 0$ $3x^2-4x+4!=0$

$discriminant < 0$ $"discriminant"<0$

$=>I_1=RR$

$x^2-3>=0$

$(x-3)(x+3)>=0$

$I_2=(-oo,-3]uuu[3,oo)$

$D(f)=I_1nnnI_2=RRnnn((-oo,-3]uuu[3,oo))$

$D(f)=(-oo,-3]uuu[3,oo)$