## Newton's Method for Finding Roots

Read this section. Work through practice problems 1-6.

Practice 1: $\quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+3 \mathrm{x}+1$ so $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+3$ and the slope of the tangent line at the point $(1,3)$ is $\mathrm{f}^{\prime}(1)$ $=6$. Using the point-slope form for the equation of a line, the equation of the tangent line is $y-3=6(x-1)$ or $y=6 x-3$.

The $\mathrm{y}$-coordinate of a point on the $\mathrm{x}$-axis is 0 so we need to put $\mathrm{y}=0$ and solve the linear equation for $\mathrm{x}: 0=6 \mathrm{x}-3$ so $\mathrm{x}=1 / 2$.

The line tangent to the graph of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+3 \mathrm{x}+1$ at the point $(1,3)$ intersects the $\mathrm{x}$-axis at the point $(\mathbf{1} / \mathbf{2}, \mathbf{0})$.

Practice 2: The approximate locations of $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ are shown in Fig. 20.

Practice 3: $\quad f(x)=x^{3}-3 x^{2}+x-1$ so $f^{\prime}(x)=3 x^{2}-6 x+1. x_{0}=3$.

\begin{aligned} &\mathrm{x}_{1}=\mathrm{x}_{0}-\frac{\mathrm{f}\left(\mathrm{x}_{0}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)}=3-\frac{\mathrm{f}(3)}{\mathrm{f}^{\prime}(3)}=3-\frac{2}{10}=\mathbf{2. 8} \\ &\mathrm{x}_{2}=\mathrm{x}_{1}-\frac{\mathrm{f}\left(\mathrm{x}_{1}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{1}\right)}=2.8-\frac{\mathrm{f}(2.8)}{\mathrm{f}^{\prime}(2.8)}=2.8-\frac{0.232}{7.72} \approx 2.769948187 \\ &\mathrm{x}_{3}=\mathrm{x}_{2}-\frac{\mathrm{f}\left(\mathrm{x}_{2}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{2}\right)} \approx 2.769292663 \end{aligned}

Practice 4: Fig. 21 shows the first iteration of Newton's Method for $x_{0}=2,3$, and $\mathrm{5}$.

If $\mathrm{x}_{0}=2$, the iterates approach the root at $\mathrm{a}$.

If $x_{0}=3$, the iterates approach the root at $\mathrm{c}$.

If $x_{0}=5$, the iterates approach the root at $\mathrm{a}$.

Practice 5: $\quad f(x)=x^{1 / 3}$ so $f^{\prime}(x)=\frac{1}{3} x^{-2 / 3}$.

If $x_{0}=1$, then $\quad x_{1}=1-\frac{f(1)}{f^{\prime}(1)}=1-\frac{1}{1 / 3}=1-3=-2$

$x_{2}=-2-\frac{\mathrm{f}(-2)}{\mathrm{f}^{\prime}(-2)}=-2-\frac{(-2)^{1 / 3}}{\frac{1}{3}(-2)^{-2 / 3}}=-2-\frac{-2}{1 / 3}=4$

$\mathrm{x}_{3}=4-\frac{\mathrm{f}(4)}{\mathrm{f}^{\prime}(4)}=4-\frac{(4)^{1 / 3}}{\frac{1}{3}(4)^{-2 / 3}}=4-\frac{4}{1 / 3}=-8$, and so on

If $x_{0}=-3$, then $x_{1}=-3-\frac{f(-3)}{f^{\prime}(-3)}=-3-\frac{(-3)^{1 / 3}}{\frac{1}{3}(-3)^{-2 / 3}}=-3+9=6$

$x_{2}=6-\frac{f(6)}{f^{\prime}(6)}=6-\frac{6^{1 / 3}}{\frac{1}{3} 6^{-2 / 3}}=6-\frac{6}{1 / 3}=-12$

The graph of the cube root $\mathrm{f}(\mathrm{x})=\mathrm{x}^{1 / 3}$ has a shape similar to Fig. 14, and the behavior of the iterates is similar to the pattern in that figure. Unless $\mathrm{x}_{0}=0$ (the only root of $\mathrm{f}$) the iterates alternate in sign and double in magnitude with each iteration: they get progressively farther from the root with each iteration.

Practice 6: If $\mathrm{x}_{0}=0.997$, then $\mathrm{x}_{1} \approx-0.003, \mathrm{x}_{2} \approx 166.4, \mathrm{x}_{3} \approx 83.2, \mathrm{x}_{4} \approx 41.6$.

If $\mathrm{x}_{0}=1.02$, then $\mathrm{x}_{1} \approx 0.198, \mathrm{x}_{2} \approx-25.2376, \mathrm{x}_{3} \approx-12.6, \mathrm{x}_{4} \approx-6.26$