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

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!

• Absolutely sure that "the idea is to use Banach fixed point theorem"? – Did Apr 12 '13 at 5:36
• The chapter is 'Complete metric spaces' and it says use theorem X, which states Banach fixed point theorem but for intervals in $R$. – elaRosca Apr 12 '13 at 5:38
• I am asking this because, by the IVT, the result is a direct consequence of the fact that $f$ is increasing on $[0,1]$ with $f(0)\lt0\lt f(1)$. – Did Apr 12 '13 at 5:47
• Could you state that the Banach's fixed point theorem? – Paul Apr 12 '13 at 6:23

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