Hindi, asked by Somali253, 1 year ago

If f:[1,∞) → [2,∞) is given by f(x) = x+1/x find f^-1(x) , {assume bijection)

Answers

Answered by Skidrow
9
Let y = f(x)

therefore \: y =  \frac{ {x}^{2} + 1 }{x}  =  >  {x}^{2}  - xy + 1 = 0 \\  =  > x =  \frac {y    + _{ - }  \sqrt{ {y}^{2}  - 4}  }{2 }  \\  =  > f ^{ - 1} (y) =  \frac{y  + _{ - } \sqrt{ {y}^{2} - 4 }  }{2}  \:  \: \: ( as \: f(x) = y = >  x = f ^{ - 1} (y) \\  \\  =  > f ^{ - 1}(x) =  \frac{x  + _{ - } \sqrt{ {x}^{2} - 4 }  }{2}

Since the range of inverse function is [1,∞), therefore neglecting the negative sign, we have

f ^{ - 1} (x) =  \frac{x +  \sqrt{ {x}^{2} - 4 } }{2}
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