$11 f(x)$ is a function defined on $\mathbf{R}$, and it is an even function. Moreover, $f(1+x)=f(1-x)$, and $f(x)$ is monotonically increasing in the interval $(0,1)$. Then, the size relationship between $f(8.5)$ and $f(-4.8)$ is $\qquad$
(11) $f(8.5)<f(-4.8)$ Hint: Since $f(x)$ is an even function, and $f(1+x)=f(1-x)$, therefore
$$
f(1+x)=f(1-x)=f(x-1) .
$$

Let $x-1=t$, we get $f(t+2)=f(t)$, which means $f(x)$ is an even function with a period of 2.
$$
\begin{array}{c}
f(8.5)=f(2 \times 4+0.5)=f(0.5), \\
f(-4.8)=f(4.8)=f(2 \times 2+0.8)=f(0.8) .
\end{array}
$$

Since $f(x)$ is monotonically increasing in the interval $(0,1)$, we have $f(0.5)<f(0.8)$, thus $f(8.5)<f(-4.8)$.