Find the domain and range of 1/ root of 16-x2

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Solution

$D e a r S t u d e n t, D o m a i n : (- 4, 4) S o l u t i o n - l e t f (x) = \frac{1}{\sqrt{16 - x^{2}}} f o r f u n c t i o n t o b e d e f i n e d; 16 - x^{2} > 0 4^{2} > x^{2} 4 > | x | - 4 < x < 4 R a n g e : [\frac{1}{4}, \infty) S o l u t i o n : 0 < \sqrt{16 - x^{2}} \leq \sqrt{16} 0 < \sqrt{16 - x^{2}} \leq 4 \frac{1}{4} \leq \frac{1}{\sqrt{16 - x^{2}}} < \infty R e g a r d s$

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