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!
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 $1$.