Is f a Contraction Mapping on [1,∞)?

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SUMMARY

The function f defined as f(x) = x^{0.5} + x^{-0.5} on the interval [1, ∞) is analyzed for contraction mapping properties. The Mean Value Theorem (MVT) was initially considered but proved ineffective for this function. The analysis reveals that for 1 ≤ x < y < +∞, the difference |f(y) - f(x)| can be expressed as (1/2√c)(1 - 1/c)|y - x|, where c is a value in the interval. The global maximum for F(c) = (1/2√c)(1 - 1/c) is determined to be K = 1/(3√3), which is less than 1, confirming that f is indeed a contraction mapping on the specified interval.

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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|
 
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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|

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 $K=\dfrac{1}{3\sqrt{3}}<1$
 

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