Answer
$3.22857729$
Work Step by Step
Step 1. Given $f(x)=x^4-x^3$, let $f(x)=75$, we have the equation $x^4-x^3=75$ or $g(x)=x^4-x^3-75=0$
Step 2. Testing end point values of the interval $[3,4]$, we have $g(3)=3^4-3^3-75=-21\lt0$ and $g(4)=4^4-4^3-75=117\gt0$. As $g(x)$ is a continuous function over the interval, using the Intermediate Value Theorem, function $g(x)$ has a zero in the interval $[3,4]$
Step 3. Find $g'(x)=4x^3-3x^2$ and use Newton’s method $x_{n+1}=x_n-\frac{g(x_n)}{g'(x_n)}$ with a starting point $x_0=3$. We can obtain the zero shown in the table as $3.22857729$
