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If (f:[3,\infty)\to[1,\infty)) is defined by (f(x)=x^2-6x+10), what is (f^{-1}(x))?यदि (f:[3,\infty)\to[1,\infty)) को (f(x)=x^2-6x+10) से परिभाषित किया गया है, तो (f^{-1}(x)) क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (3+\sqrt{x-1})(3+\sqrt{x-1})
Step 1: (f(x)=x^2-6x+10=(x-3)^2+1). Step 2: From (y=(x-3)^2+1), we get ((x-3)^2=y-1). Step 3: Since the domain is ([3,\infty)), (x-3\geq0), so the inverse is (3+\sqrt{x-1}).चरण 1: (f(x)=x^2-6x+10=(x-3)^2+1)। चरण 2: (y=(x-3)^2+1) से ((x-3)^2=y-1) मिलता है। चरण 3: प्रांत ([3,\infty)) है, इसलिए (x-3\geq0) और प्रतिलोम (3+\sqrt{x-1}) होगा।
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