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Question Number 107674 by Study last updated on 12/Aug/20

f(x)=(√x)(√x)       D_f =???

$${f}\left({x}\right)=\sqrt{{x}}\sqrt{{x}}\:\:\:\:\:\:\:{D}_{{f}} =??? \\ $$

Answered by Her_Majesty last updated on 12/Aug/20

x=re^(iθ) ; r∈R^+ ∧θ∈R  ⇒  f(x)=(√(re^(iθ) ))(√(re^(iθ) ))=(√r)e^(iθ/2) (√r)e^(iθ/2) =re^(iθ) =x

$${x}={re}^{{i}\theta} ;\:{r}\in\mathbb{R}^{+} \wedge\theta\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)=\sqrt{{re}^{{i}\theta} }\sqrt{{re}^{{i}\theta} }=\sqrt{{r}}{e}^{{i}\theta/\mathrm{2}} \sqrt{{r}}{e}^{{i}\theta/\mathrm{2}} ={re}^{{i}\theta} ={x} \\ $$

Answered by Her_Majesty last updated on 12/Aug/20

x≥0 ⇒ f(x)=x  x<0 ⇒ f(x)=i(√(−x))×i(√(−x))=−1(−x)=x  ⇒ f(x)=x

$${x}\geqslant\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)={x} \\ $$$${x}<\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)={i}\sqrt{−{x}}×{i}\sqrt{−{x}}=−\mathrm{1}\left(−{x}\right)={x} \\ $$$$\Rightarrow\:{f}\left({x}\right)={x} \\ $$

Commented by Study last updated on 12/Aug/20

what is the domain of f(x)?

$${what}\:{is}\:{the}\:{domain}\:{of}\:{f}\left({x}\right)? \\ $$

Commented by Her_Majesty last updated on 12/Aug/20

if we stay in R the domain is x≥0

$${if}\:{we}\:{stay}\:{in}\:\mathbb{R}\:{the}\:{domain}\:{is}\:{x}\geqslant\mathrm{0} \\ $$

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