Question

Consider a nonlinear algebraic equation, $x\ln x + x - 1 = 0$. Using the Newton–Raphson method, with the initial guess of $x_{0}=3$, the value of $x$ after one iteration (rounded off to one decimal place) is _________.

Hint:

Newton–Raphson method converges rapidly if the initial guess is close to the actual root.

Answer 1

Correct Answer: 1.2

Given:
\[ f(x) = x \ln x + x - 1 \]
Derivative:
\[ f'(x) = \ln x + 1 + 1 = \ln x + 2 \]
Newton–Raphson formula:
\[ x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})} \]
Compute at \(x_{0} = 3\):
\[ f(3) = 3\ln 3 + 3 - 1 \]
\[ \ln 3 = 1.0986 \]
\[ f(3) = 3(1.0986) + 2 = 3.2958 + 2 = 5.2958 \]
Now compute derivative:
\[ f'(3) = \ln 3 + 2 = 1.0986 + 2 = 3.0986 \]
Apply Newton–Raphson:
\[ x_{1} = 3 - \frac{5.2958}{3.0986} \]
\[ x_{1} = 3 - 1.708 \approx 1.292 \]
Rounded to one decimal place:
\[ \boxed{1.3} \]