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Question Number 68546 by aliesam last updated on 13/Sep/19

find the range and domain of f(x)    f(x)=(√(sin^(−1) (ln(x/(10)))))

$${find}\:{the}\:{range}\:{and}\:{domain}\:{of}\:{f}\left({x}\right) \\ $$$$ \\ $$$${f}\left({x}\right)=\sqrt{{sin}^{−\mathrm{1}} \left({ln}\frac{{x}}{\mathrm{10}}\right)} \\ $$

Answered by MJS last updated on 13/Sep/19

(√u)∈R ⇒ u≥0  arcsin v ∈R ⇒ −1≤v≤1  (√(arcsin v))∈R ⇒ 0≤v≤1  0≤ln (x/(10)) ≤1 ⇒ 10≤x≤10e  domain: 10≤x≤10e  range: 0≤y≤(π/(√2))

$$\sqrt{{u}}\in\mathbb{R}\:\Rightarrow\:{u}\geqslant\mathrm{0} \\ $$$$\mathrm{arcsin}\:{v}\:\in\mathbb{R}\:\Rightarrow\:−\mathrm{1}\leqslant{v}\leqslant\mathrm{1} \\ $$$$\sqrt{\mathrm{arcsin}\:{v}}\in\mathbb{R}\:\Rightarrow\:\mathrm{0}\leqslant{v}\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant\mathrm{ln}\:\frac{{x}}{\mathrm{10}}\:\leqslant\mathrm{1}\:\Rightarrow\:\mathrm{10}\leqslant{x}\leqslant\mathrm{10e} \\ $$$$\mathrm{domain}:\:\mathrm{10}\leqslant{x}\leqslant\mathrm{10e} \\ $$$$\mathrm{range}:\:\mathrm{0}\leqslant{y}\leqslant\frac{\pi}{\sqrt{\mathrm{2}}} \\ $$

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