# Show root using Banach Fixpoint

I'm required to show that: $$f(x) = e^x - 4x$$ has a root in $$(0,1)$$ using the Banach Fixpoint theorem.

The fact that $$f((0,1)) \neq (0,1)$$ confuses me. How do I proceed without knowing that $$f$$ isn't a map of a set onto itself?

• You may try some curve sketching and find a domain so that $f(D) \subseteq D$. – GNUSupporter 8964民主女神 地下教會 Feb 12 at 15:58
• @GNUSupporter8964民主女神地下教會 How would I go about finding such a domain $D$? – user7802048 Feb 12 at 16:11
• Sorry. I think my previous comment isn't correct. Perhaps we need to divide $f$ by a certain constant, so that $f$ is a contraction. – GNUSupporter 8964民主女神 地下教會 Feb 12 at 16:17

## 1 Answer

You have to reformulate the requirement $$f(x^\star) = e^{x^\star} -4x^\star = 0$$ for a root such that the root $$x^\star$$ is a fixpoint of a function $$g(x)$$, that is $$g(x^\star) = x^\star$$. This is the case for $$g(x) = f(x)/4 +x = e^x/4$$.

Now you have to show that $$g(0,1) \subset (0,1)$$ and $$|g'(x)|<1$$ on $$x \in (0,1)$$. Then, there is a fixpoint in $$(0,1).$$