Question

Find the value of the cube root of the number $45$. (Round off your number to the nearest whole number)

Answer 1

Find the cube root

Let $x^3=45$

Therefore $f\left(x\right)=x^3-45$

From Newton Raphson method we know that

$f\left(x_{n+1}\right)=x_n-\frac{f\left(x_n\right)}{f\text{'}\left(x_n\right)}$

We know $3^3$ is $27<45$

So the first approximation is $x_0=3$

$$\begin{gathered} \therefore f\left(x_1\right)=x_0-\frac{x_0^3-45}{3x_0^2} \\ =3-\frac{27-45}{27} \\ =3.66 \end{gathered}$$

So the nearest whole number is $4$

Hence, the required final answer is .$4$.