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Show that the contraction $T(x)= (1+x)^{1/3} $ on the interval $I=[1,2]$ satisfies the definition of a contraction.

It's not just this problem-- on this site and others explanations will say "the Mean Value Theorem will show" that a constant $\alpha \in (0,1)$ satisfies the definition of a contraction on the interval. For this particular example I even know the answer is $\alpha = \frac{2^{1/3}}{6}$. My understanding of the MVT is a specific value $c$, not bounded values.

This'll probably be a quick answer, but maybe in the future others can reference it for assistance, as well!

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The MVT by itself will not give the value of the contraction constant. However, it can be used to show that such a constant indeed exists (this is all the problem requires) and also give an upper bound for it. – David Mitra May 1 at 17:02

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Let $1 \leq x_1 < x_2 \leq 2$. By the Mean Value Theorem, $T(x_2)-T(x_1) = T'(c)(x_2-x_1)$ for some $c$ strictly between $x_1$ and $x_2$, so $|T(x_2)-T(x_1)| = |T'(c)||x_2-x_1|$. Now compute $T'(x)$ and find an upper bound on $|T'(x)|$ for $x \in [1,2]$ that is less than $1$.

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Thanks! Is it just then a standard calculus exercise to calculate the (local) maximum of the function, and restrict it to the interval we're working with? – Questioneer May 1 at 18:15
@Questioneer : yes, and sometimes it is even easier than that. You just have to find $\lambda < 1$ with $|T'(x)| \leq \lambda$ for all $x \in [1,2]$, you don't even have to find $\max\{|T'(x)| \mid 1 \leq 2\}$. In this problem, I think $T'(x)$ is monotone and doesn't change sign, so finding $\max\{|T'(x)| \mid 1 \leq 2\}$ should be easy. – Stefan Smith May 1 at 20:52
@Questioneer : my previous comment is too strong. You just have to show there exists $\lambda < 1$ with $|T'(x)| \leq \lambda$ for all $x$ in the interval, you don't even have to find what $\lambda$ is. – Stefan Smith May 5 at 16:03

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