Question Number 125814 by Tanuidesire last updated on 14/Dec/20
![Let f:[1,5]→R be defined by f(x)=(6/(x+1)). Show that f has a unique fixed point and find it.](Q125814.png)
$${Let}\:{f}:\left[\mathrm{1},\mathrm{5}\right]\rightarrow\mathbb{R}\:{be}\:{defined}\:{by}\:{f}\left({x}\right)=\frac{\mathrm{6}}{{x}+\mathrm{1}}.\:{Show}\:{that}\:{f}\:{has}\:{a}\:{unique}\:{fixed}\:{point}\:{and}\:{find}\:{it}. \\ $$
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 }.](Q125860.png)
$${f}\:\mathrm{strictly}\:\mathrm{decreases}\:\mathrm{and}\:: \\ $$$${f}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{6}}{\mathrm{1}+\mathrm{1}}\:=\:\mathrm{3} \\ $$$${f}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{6}}{\mathrm{5}+\mathrm{1}}\:=\:\mathrm{1} \\ $$$${f}\left(\left[\mathrm{1},\mathrm{5}\right]\right)\:=\:\left[\mathrm{1},\mathrm{3}\right]\:\subseteq\:\left[\mathrm{1},\mathrm{5}\right] \\ $$$$\mathrm{and}\:: \\ $$$${f}\left({x}\right)\:=\:{x}\:\Leftrightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{6}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:{x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{6}\right)}}{\mathrm{2}}\:=\:−\mathrm{3}\:\mathrm{or}\:\mathrm{2} \\ $$$$\mathrm{Only}\:{x}\:=\:\mathrm{2}\:\mathrm{is}\:\mathrm{included}\:\mathrm{in}\:\left[\mathrm{1},\mathrm{5}\right]. \\ $$$$\Rightarrow\:\mathcal{S}\:=\:\left\{\:\mathrm{2}\:\right\}. \\ $$