Answer
$2.94283095$
Work Step by Step
First, create a function based on $\sqrt[4] {75}$ : $f(x) = x^{4} -75$
Next, find its derivative : $f'(x) = 4x^{3}$
Now, we do know that $\sqrt[4] {16} < \sqrt[4] {75} < \sqrt[4] {81}$, so we will pick a number between $2$ and $3$. Pick $x_1 = 2.5$.
Using the Newton's formula, we get:
$x_2 = x_1 - \frac{f(x)}{f'(x)} = 2.5 - \frac{2.5^4 - 75}{4\cdot2.5^{3}} = 3.075$
$x_3 = 3.075 - \frac{3.075^4-75}{4\cdot3.075^3} = 2.95111107$
$x_4 = 2.951111 - \frac{2.95111107^4-75}{4\cdot 2.95111107^3} = 2.94286573$
$x_5 = 2.942865 - \frac{2.9428673^4-75}{4\cdot2.94286573^3} = 2.94283095$
$x_6 = 2.942830 - \frac{2.94283095^4-75}{4\cdot2.94283095^3} = 2.94283095$
We see that $x_5 = x_6$, which means that we approximated correct to eight decimal places.
Therefore, the approximation up to 8 decimal places of $\sqrt[4] {75}$ is $2.94283095$.
