Suppose that Newton's method is used to locate a root of the equation f (x) = 0 with initial approximation x1 = 7. If the second approximation is found to be x2 = −8, and the tangent line to f (x) at x = 7 passes through the point (10, 7), find f (7).
Suppose that Newton's method is used to locate a root of the equation f (x) = 0 with initial approximation x1 = 7.
If the second approximation is found to be x2 = −8,
and the tangent line to f (x) at x = 7 passes through the point (10, 7),
find f (7).
Newton's method:
\(\begin{array}{|rcll|} \hline x_1 - x_2 &=& \dfrac{f(x_1)}{f'(x_1)} \quad & | \quad x_1 = 7 \quad x_2 = -8 \\\\ 7+8 &=& \dfrac{f(7)}{f'(7)} \\\\ 15 &=& \dfrac{f(7)}{f'(7)} \\\\ f(7) &=& 15f'(7) \quad & | \quad f'(7)=\dfrac{7}{10} \\ f(7) &=& 15\cdot \dfrac{7}{10} \\ \mathbf{f(7)} &\mathbf{=}& \mathbf{10.5} \\ \hline \end{array} \)
Suppose that Newton's method is used to locate a root of the equation f (x) = 0 with initial approximation x1 = 7.
If the second approximation is found to be x2 = −8,
and the tangent line to f (x) at x = 7 passes through the point (10, 7),
find f (7).
Newton's method:
\(\begin{array}{|rcll|} \hline x_1 - x_2 &=& \dfrac{f(x_1)}{f'(x_1)} \quad & | \quad x_1 = 7 \quad x_2 = -8 \\\\ 7+8 &=& \dfrac{f(7)}{f'(7)} \\\\ 15 &=& \dfrac{f(7)}{f'(7)} \\\\ f(7) &=& 15f'(7) \quad & | \quad f'(7)=\dfrac{7}{10} \\ f(7) &=& 15\cdot \dfrac{7}{10} \\ \mathbf{f(7)} &\mathbf{=}& \mathbf{10.5} \\ \hline \end{array} \)