Unique solution to $x^4 + 7x -1 = 0$ on $[0,1]$ (Banach's fixed point theorem)

Question

I want to show that $x^4 + 7x -1 = 0$ has a unique solution on $[0,1]$.

The idea is to use Banach's fixed point theorem. However, I see a problem with this as the statement of the theorem says that the function $f$ has to be defined from a complete metric space to itself. While $[0,1]$ is complete (as a closed subset of $\mathbb{R}$ which is complete) the function $f(x) = x ^ 4 + 8x - 1 $ does not have an image only in $[0,1]$.

What am I missing/not understanding here?

Thanks a lot!

Answer 1

Try to apply Banach fixed-point theorem to the function $$f(x)=\frac{1-x^4}7$$ on the interval $[0,1]$.