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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).

 Sep 20, 2018

Best Answer 

 #1
avatar+21978 
+10

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:

\(\begin{array}{|rcll|} \hline x_1 - x_2 &=& \dfrac{f(x_1)}{f'(x_1)} \quad & | \quad x_1 = 7 \quad x_2 = -8 \\\\ 7+8 &=& \dfrac{f(7)}{f'(7)} \\\\ 15 &=& \dfrac{f(7)}{f'(7)} \\\\ f(7) &=& 15f'(7) \quad & | \quad f'(7)=\dfrac{7}{10} \\ f(7) &=& 15\cdot \dfrac{7}{10} \\ \mathbf{f(7)} &\mathbf{=}& \mathbf{10.5} \\ \hline \end{array} \)

 

laugh

 Sep 20, 2018
 #1
avatar+21978 
+10
Best Answer

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:

\(\begin{array}{|rcll|} \hline x_1 - x_2 &=& \dfrac{f(x_1)}{f'(x_1)} \quad & | \quad x_1 = 7 \quad x_2 = -8 \\\\ 7+8 &=& \dfrac{f(7)}{f'(7)} \\\\ 15 &=& \dfrac{f(7)}{f'(7)} \\\\ f(7) &=& 15f'(7) \quad & | \quad f'(7)=\dfrac{7}{10} \\ f(7) &=& 15\cdot \dfrac{7}{10} \\ \mathbf{f(7)} &\mathbf{=}& \mathbf{10.5} \\ \hline \end{array} \)

 

laugh

heureka Sep 20, 2018
 #2
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The tangent line at x = 7 passes through the points (7, f(7)) and (10, 7),

in which case

\(\displaystyle f\;'(7) = \frac{7-f(7)}{10-7}.\)

.
 Sep 20, 2018

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