Answer
$x=1.15$
Work Step by Step
We test $f(x)=x^{3}+x^{2}+x-4$
on nonnegative integer values of x, and find that
$f(1) $is negative, $f(2)$is positive $\Rightarrow$ by the IVT, a real zero is inside $[1,2].$
Subdividing $[1,2]$ into 10 subintervals, calculating f( endpoints), we find
$f(1.1) $is negative, $f(1.2)$is positive $\Rightarrow$ by the IVT, a real zero is inside $[1.1,1.2].$
Subdividing again, we find
$f(1.15) $is negative, $f(1.16)$is positive $\Rightarrow$ by the IVT, a real zero is inside $[1.15,1.16].$
The real zero is $x=1.15...$
To two decimal places, $x=1.15$
