Answer
$x=0.70$
Work Step by Step
We test $f(x)=x^{3}+x^{2}+x-4$
on nonnegative integer values of x, and find that
$f(0) $ is negative, $f( 1)$ is positive $\Rightarrow$ by the IVT, a real zero is inside $[0,1].$
Subdividing $[0,1]$ into 10 subintervals, calculating f( endpoints), we find
$f(0.7) $ is negative, $f(0.8)$ is positive $\Rightarrow$ by the IVT, a real zero is inside $[0,7,0.8].$
Subdividing again, we find
$f(0.70) $ is negative, $f(0.71)$ is positive $\Rightarrow$ by the IVT, a real zero is inside $[0.70,0.71].$
The real zero is $x=0.70...$
To two decimal places, $x=0.70$
