Let f:[1,5]→R be defined by f(x)=(6/(x+1)). Show that f has a unique fixed point and find it.


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Question Number 125814 by Tanuidesire last updated on 14/Dec/20

$L e t f : [1, 5] \to R b e d e f i n e d b y f (x) = \frac{6}{x + 1} . S h o w t h a t f h a s a u n i q u e f i x e d p o i n t a n d f i n d i t .$
	Answered by Olaf last updated on 14/Dec/20
	$f strictly decreases and : f(1) = (6/(1+1)) = 3 f(5) = (6/(5+1)) = 1 f([1,5]) = [1,3] ⊆ [1,5] and : f(x) = x ⇔ x^2 +x−6 = 0 ⇔ x = ((−1±(√(1−4(1)(−6))))/2) = −3 or 2 Only x = 2 is included in [1,5]. ⇒ S = { 2 }.$
	$f strictly decreases and :$ $f (1) = \frac{6}{1 + 1} = 3$ $f (5) = \frac{6}{5 + 1} = 1$ $f ([1, 5]) = [1, 3] \subseteq [1, 5]$ $and :$ $f (x) = x \Leftrightarrow x^{2} + x - 6 = 0$ $\Leftrightarrow x = \frac{- 1 \pm \sqrt{1 - 4 (1) (- 6)}}{2} = - 3 or 2$ $Only x = 2 is included in [1, 5] .$ $\Rightarrow S = {2} .$
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