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If \(f(x)=\sqrt{x+2}\) and \(g(x)=\sqrt{6-x}\), what is the real domain of \(\left(\frac{f-g}{f+g}\right)(x)\)?यदि \(f(x)=\sqrt{x+2}\) और \(g(x)=\sqrt{6-x}\) हैं, तो \(\left(\frac{f-g}{f+g}\right)(x)\) का वास्तविक प्रांत क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([-2,6])([-2,6])
For both square roots, \(x+2\ge0\) and \(6-x\ge0\), so \(-2\le x\le6\). The denominator (f(x)+g(x)) is never (0) on this interval, so no point is removed.दोनों वर्गमूलों के लिए \(x+2\ge0\) और \(6-x\ge0\), इसलिए \(-2\le x\le6\)। हर (f(x)+g(x)) इस अंतराल में (0) नहीं होता, इसलिए कोई बिंदु हटता नहीं है।
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