Jan 16, 2013 Thread starter Banned #1 P Poirot Banned Feb 15, 2012 250 f:[1,infinity)->[1,infinity) $f(x)=x^{0.5}+x^{-0.5}$ I thought about using MVT but it doesn't work and I've tried showing it conventially but i can't reduce it to k|x-y|
f:[1,infinity)->[1,infinity) $f(x)=x^{0.5}+x^{-0.5}$ I thought about using MVT but it doesn't work and I've tried showing it conventially but i can't reduce it to k|x-y|
Jan 16, 2013 #2 Fernando Revilla Well-known member MHB Math Helper Jan 29, 2012 661 Poirot said: f:[1,infinity)->[1,infinity) $f(x)=x^{0.5}+x^{-0.5}$ I thought about using MVT but it doesn't work and I've tried showing it conventially but i can't reduce it to k|x-y| Click to expand... For $1\leq x <y<+\infty$ you'll get $\left|f(y)-f(x)\right|=\dfrac{1}{2\sqrt{c}}\left(1-\dfrac{1}{c}\right)|y-x|$ Now, use that a global maximum for $F(c)=\dfrac{1}{2\sqrt{c}}\left(1-\dfrac{1}{c}\right)$ in $[1,+\infty)$ is $K=\dfrac{1}{3\sqrt{3}}<1$
Poirot said: f:[1,infinity)->[1,infinity) $f(x)=x^{0.5}+x^{-0.5}$ I thought about using MVT but it doesn't work and I've tried showing it conventially but i can't reduce it to k|x-y| Click to expand... For $1\leq x <y<+\infty$ you'll get $\left|f(y)-f(x)\right|=\dfrac{1}{2\sqrt{c}}\left(1-\dfrac{1}{c}\right)|y-x|$ Now, use that a global maximum for $F(c)=\dfrac{1}{2\sqrt{c}}\left(1-\dfrac{1}{c}\right)$ in $[1,+\infty)$ is $K=\dfrac{1}{3\sqrt{3}}<1$