Answer
$1.17951$
Work Step by Step
Step 1. Given $f(x)=x^3+2x-4$, we have $f'(x)=3x^2+2$.
Step 2. Test signs of $f(x)$, $f(1)=1+2-4=-1\lt0$ and $f(2)=8+4-4=8\gt0$; thus there is a zero in the interval of $(1,2)$ based on the Intermediate Value Theorem of a continuous function.
Step 3. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=1$, we can obtain the zero shown in the table as $1.179509025\approx1.17951$
