When rounded to six decimal places, we have

x$4=$ x$3=0\mathrm{.}860334.$

Therefore, this is a six$-$decimal approximation of the critical value of

f$($x$)$

on the interval

$0$ x $.$

We now wish to find the value of f at this critical number, rounding the final result to six decimal places.

f$($x$)=11$x cos$($x$)$f$(0\mathrm{.}860334)=$