Answer
$2.195823345$
Work Step by Step
Step 1. Given $f(x)=3x-x^3$, let $f(x)=-4$, we have the equation $3x-x^3=-4$ or $g(x)=x^3-3x-4=0$
Step 2. Testing end point values of the interval $[2,3]$, we have $g(2)=8-6-4=-2\lt0$ and $g(3)=27-9-4=14\gt0$. As $g(x)$ is a continuous function over the interval, using the Intermediate Value Theorem, function $g(x)$ has a zero in the interval $[2,3]$
Step 3. Find $g'(x)=3x^2-3$ and use Newton’s method $x_{n+1}=x_n-\frac{g(x_n)}{g'(x_n)}$ with a starting point $x_0=2$. We can obtain the zero shown in the table as $2.195823345$