The approximate solution is 3.52138 (Correct choice: D)
How to find a point of intersection by Newton-Raphson method
Since [tex]f(x)[/tex] and [tex]g(x)[/tex] are continuous differentiable functions, we can apply Newton-Raphson formula, which helps to find an approximate solution by a multi-stage approach:
[tex]x_{i+1} = x_{i}-\frac{f(x_{i})}{f'(x_{i})}[/tex], where [tex]f(x_{i}) = 3\cdot \log (x-2) - \log x[/tex]. (1, 2)
The first derivative evaluated at [tex]x_{i}[/tex] is defined by this formula:
[tex]f'(x_{i}) = \frac{1.303}{x-2} - \frac{0.434}{x}[/tex] (3)
Now we proceed to find an approximate solution by iterations:
Iteration 1
[tex]x_{o} = 3.5[/tex]
[tex]f(x_{o}) = -0.016[/tex]
[tex]f'(x_{o}) = 0.745[/tex]
[tex]x_{1} = 3.521[/tex]
Iteration 2
[tex]x_{1} = 3.521[/tex]
[tex]f(x_{1}) = -2.784\times 10^{-4}[/tex]
[tex]f'(x_{1}) = 0.733[/tex]
[tex]x_{2} = 3.52137[/tex]
Iteration 3
[tex]x_{2} = 3.52137[/tex]
[tex]f(x_{2}) = -7.116\times 10^{-6}[/tex]
[tex]f'(x_{2}) = 0.7332[/tex]
[tex]x_{3} \approx 3.52138[/tex]
The approximate solution is 3.52138 (Correct choice: D) [tex]\blacksquare[/tex]
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