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What is the range of \(f(x)=\frac{x}{\sqrt{x^2+9}}\)?फलन \(f(x)=\frac{x}{\sqrt{x^2+9}}\) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( (-1,1) )( (-1,1) )
The denominator is always greater than \(|x|\), so \(\left|\frac{x}{\sqrt{x^2+9}}\right|<1\). For large \(|x|\), the value approaches \(\pm 1\) but never equals them.हर हमेशा \(|x|\) से बड़ा है, इसलिए \(\left|\frac{x}{\sqrt{x^2+9}}\right|<1\)। बड़े \(|x|\) पर मान \(\pm 1\) के पास जाता है पर बराबर नहीं होता।
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