Show using the fixed point theorem that if $f(x)=\frac{1}{4}[1-x-\frac{1}{10}x^5]$ is defined on $I=\{x|0 \leq x \leq 1\}$ then it has a zero in $I$

Question

The main idea here is to apply the fixed point theorem to $g(x)=f(x)+x$, in order to show that f has a zero in $I$. If $g$ has a fixed point (i.e. $g(x_0)=x_0$), then $f(x_0)=0$. I just don't see how to show that $d(g(x),g(y))=d(f(x)+x,f(y)+y) \leq kd(x,y)$ for all $x,y \in I$, and with $0<k<1$, which is necessary to apply the fixed point theorem to $g$.

EDIT:

This is an exercise designed to help me learn how to use the following fixed point theorem:

If $f$ is a mapping on a complete metric space $S$ into $S$ such that $d(f(x),f(y)) \leq kd(x,y)$ for all $x,y \in S$ with $0<k<1$, then the mapping has a unique fixed point.

Answer 1

By the MVT, $g(x)-g(y) =g'(\xi) (x-y)$ for $0<\xi<1$. Thus,

$$\begin{align} |g(x)-g(y)| & =|g'(\xi)| |(x-y)| \\ & = \left(\frac34 -\frac18 \xi^4\right) |x-y| \\ & \le \frac34 |x-y| \\ & < (1) |x-y| \end{align}$$

Answer 2

$$\begin{align}|g(x)-g(y)|&=\left|\frac{3}{4}(x-y)-\frac{1}{40}(x^5-y^5)\right|\\ &\leq\frac{3}{4}\left|x-y\right|+\frac{1}{40}\left|x^5-y^5\right|\\ &=\frac{3}{4}\left|x-y\right|+\frac{1}{40}\left|x-y\right|(x^4+x^3y+x^2y^2+xy^3+y^4)\\ &\leq\frac{3}{4}\left|x-y\right|+\frac{5}{40}\left|x-y\right|\\ &=\frac{35}{40}\left|x-y\right|\end{align}$$

You can take $k=35/40 = 7/8<1$.