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Question Number 117934 by Ar Brandon last updated on 14/Oct/20

Let f : [1, \infty) \to [2, \infty) be the function defined by

f (x) = x + \frac{1}{x}

If g : [2, \infty) \to [1, \infty), is a function such that (g \circ f) (x) = x

for all x ⩾ 1 . Show that g (t) = \frac{t + \sqrt{t^{2} - 4}}{2}

Answered by Lordose last updated on 14/Oct/20

g (f (x)) = x

g (x + \frac{1}{x}) = x

x + \frac{1}{x} = y

x^{2} - yx + 1 = 0

x^{2} - yx + {(- \frac{y}{2})}^{2} = - 1 + \frac{y^{2}}{4}

(x - \frac{y}{2}) = \pm \sqrt{\frac{y^{2} - 4}{4}}

x = \frac{1}{2} (y \pm \sqrt{y^{2} - 4})

g (y) = \frac{1}{2} (y \pm \sqrt{y^{2} - 4})

g (t) = \frac{1}{2} (t \pm \sqrt{t^{2} - 4})

Commented by Ar Brandon last updated on 14/Oct/20

Thank You ��