Question

Using Newton-Raphson method, the cube root of $24$ is?

• A. $2.884$
• B. $3.256$
• C. $5.231$
• D. $4.526$

Answer 1

The correct option is A $2.884$

To find the cube root of $24$ using Newton - Raphson method,
we need to solve $f\left(x\right)=x^3-24$.
$\Rightarrow f^{\prime }\left(x\right)=3x^2$
Notice $3^3=27$
Therefore the cube root of $24$ is slightly less than $3$.
We have $f\left(x\right)=x^3-24,f^{\prime }\left(x\right)=3x^2$
Let us start estimating the root $x$
Let the first estimation be $a=2.9$ (slightly less than 3)
Hence the subsequent estimates will be $b=a-\frac{f\left(a\right)}{f^{\prime }\left(a\right)},c=b-\frac{f\left(b\right)}{f^{\prime }\left(b\right)}$.
$f\left(a\right)=f\left(2.9\right)=(2.9{)}^3-24=0.389$ and $f^{\prime }\left(a\right)=f^{\prime }\left(2.9\right)=3(2.9{)}^2=25.23$
Therefore $b=2.9-\frac{0.389}{25.23}\approx 2.88458$
Now $c=2.88458-\frac{f\left(2.88458\right)}{f^{\prime }\left(2.88458\right)}=2.88449$
Hence the cube root of $24$ is $2.884$