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Felix's question at Yahoo! Answers regarding Newton's method
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Hello Felix
Since the line $y=5x-2$ is tangent to the function $f(x)$ at $x=4$, we know two things:
a) The function and the line have a common point at $x=4$:
\(\displaystyle f(4)=5(4)-2=18\)
b) At $x=4$ the function's derivative is equal to the slope of the line:
\(\displaystyle f'(4)=5\)
Now, Newton's method gives us:
\(\displaystyle x_{n+1}=x_{n}-\frac{f\left(x_n \right)}{f'\left(x_n \right)}\)
If the initial approximation is \(\displaystyle x_1=4\), then the second approximation is:
\(\displaystyle x_{2}=x_{1}-\frac{f\left(x_1 \right)}{f'\left(x_1 \right)}=4-\frac{18}{5}=\frac{2}{5}\)
As we should expect, the second guess is simply the root of the tangent line:
\(\displaystyle 0=5x_2-2\implies x_2=\frac{2}{5}\)
Since the line $y=5x-2$ is tangent to the function $f(x)$ at $x=4$, we know two things:
a) The function and the line have a common point at $x=4$:
\(\displaystyle f(4)=5(4)-2=18\)
b) At $x=4$ the function's derivative is equal to the slope of the line:
\(\displaystyle f'(4)=5\)
Now, Newton's method gives us:
\(\displaystyle x_{n+1}=x_{n}-\frac{f\left(x_n \right)}{f'\left(x_n \right)}\)
If the initial approximation is \(\displaystyle x_1=4\), then the second approximation is:
\(\displaystyle x_{2}=x_{1}-\frac{f\left(x_1 \right)}{f'\left(x_1 \right)}=4-\frac{18}{5}=\frac{2}{5}\)
As we should expect, the second guess is simply the root of the tangent line:
\(\displaystyle 0=5x_2-2\implies x_2=\frac{2}{5}\)