Question

Using Newton-Raphson method, find the cube root of 24 ?
A. 2.884
B. 3.256
C. 5.231
D. 4.526

Answer 1

Hint: According to Newton- Raphson method we can find approximate value as
 ${{x}_{n+1}}={{x}_{n}}-\dfrac{f({{x}_{n}})}{f'({{x}_{n}})}$

Complete step-by-step answer:
Let the cube root of 24 is x.
So we can write it as
$\Rightarrow x=\sqrt[3]{24}$
$\Rightarrow {{x}^{3}}=24$
$\Rightarrow {{x}^{3}}-24=0$
So $f(x)={{x}^{3}}-24$
On differentiating both side w.r.t x
$\Rightarrow f'(x)=\dfrac{d}{dx}(f(x))=\dfrac{d}{dx}\left( {{x}^{3}}-24 \right)$
$\Rightarrow f'(x)=\dfrac{d}{dx}\left( {{x}^{3}} \right)-\dfrac{d}{dx}\left( 24 \right)$
$\Rightarrow f'(x)=3{{x}^{2}}-0$ $\left\{ \because \dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}} \right\}$
$\Rightarrow f'(x)=3{{x}^{2}}$
As we know, the cube root of 27 is 3. So the cube root of 24 is slightly less than 3.
So the first estimation value is ${{x}_{1}}=2.9$ which is slightly less than 3.
According to Newton Raphson method
${{x}_{n+1}}={{x}_{n}}-\dfrac{f({{x}_{n}})}{f'({{x}_{n}})}$
For second approximation , n = 1
${{x}_{2}}={{x}_{1}}-\dfrac{f({{x}_{1}})}{f'({{x}_{1}})}$
In question ${{x}_{1}}=2.9$, $f(x)={{x}^{3}}-24$ and $f'(x)=3{{x}^{2}}$
$\Rightarrow {{x}_{2}}=2.9-\dfrac{{{(2.9)}^{3}}-24}{3\times {{(2.9)}^{2}}}$
$\Rightarrow {{x}_{2}}=2.9-\dfrac{0.389}{25.23}\approx 2.88458$
For third approximation, n = 2
 $\Rightarrow {{x}_{3}}={{x}_{2}}-\dfrac{f({{x}_{2}})}{f'({{x}_{2}})}$
$\Rightarrow {{x}_{3}}=2.88458-\dfrac{f(2.88458)}{f'(2.88458)}$
$\Rightarrow {{x}_{3}}=2.88458-\dfrac{{{(2.88458)}^{3}}-24}{3\times {{(2.88458)}^{2}}}$
$\Rightarrow {{x}_{3}}=2.88449$
As we can see in the second and third approximation value is the same till three places of decimal.
So we can say cube root of 24 is 2.884
Hence option A is correct.

Note: In Newton Raphson method if we have not given any initial approximated value then we let it according to question.
In this we continue steps until we get repeated value in two consecutive steps.