how to prove contractivity

Question

$f(x) = (x-2)^2 - ln(x)$ on the interval $[1,2]$ has a unique root? I thought it in this way: $f(x) = 0$ is equivalent to say $(x-2)^2 = ln(x)$, which is equivalent to $g(x) = exp((x-2)^2)$ has a unique fixed point, but the $L$ turns out to be that it can be greater than 1.

Answer 1

I suggest an alternative approach. We want to show that $$f(x) = (x-2)^2 - \ln(x)$$ has a unique root on $[1,2]$. We have

$$f'(x) = 2(x-2) - \frac{1}{x}$$ so clearly $f'(x) < 0$ for $x \in [1,2]$ showing that $f$ is strictly monotonically decreasing on $[1,2]$, and thus can have at most one root on $[1,2]$. Because $f(1) > 0$ and $f(2) < 0$, by the intermediate value theorem we see that $f$ in fact has a unique root in $[1,2]$.