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What is the range of (f(x)=\frac{1}{x^2+4x+8})?फलन (f(x)=\frac{1}{x^2+4x+8}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( (0,\frac{1}{4}] )( (0,\frac{1}{4}] )
The denominator ((x+2)^2+4\ge 4), so the maximum value is (\frac{1}{4}). As the denominator grows, the value approaches (0).हर ((x+2)^2+4\ge 4) है, इसलिए अधिकतम मान (\frac{1}{4}) है। हर असीम होने पर मान (0) के पास जाता है।
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