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If (f(x)=\frac{x-4}{x^2-16}), what is the real domain?यदि (f(x)=\frac{x-4}{x^2-16}), तो वास्तविक प्रान्त क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (R-{-4,4})(R-{-4,4})
Step 1: The denominator is (x^2-16=(x-4)(x+4)). Step 2: The denominator cannot be zero, so (x\ne4) and (x\ne-4). Step 3: Even if (x-4) cancels, (x=4) is not valid in the original function.चरण 1: हर (x^2-16=(x-4)(x+4)) है। चरण 2: हर शून्य नहीं हो सकता, इसलिए (x\ne4) और (x\ne-4)। चरण 3: भले ही (x-4) कटे, मूल फलन में (x=4) मान्य नहीं होगा।
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