Practice Problems

1. See Fig. 15

3. $x_{0}=1:$ a.      $x_{0}=5:$ b.

5. $\begin{aligned} &x_{0}=1: 1,2,1,2,1, \ldots \\ &x_{0}=5: x_{1} \text { is undefined since } f^{\prime}(5)=0 \end{aligned}$

7. If $\mathrm{f}$ is differentiable, then $\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)=0$ and $\mathrm{x}_{1}=\mathrm{x}_{\mathrm{o}}-\frac{\mathrm{f}\left(\mathrm{x}_{\mathrm{o}}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{\mathrm{O}}\right)}$ is undefined.

9. $f(x)=x^{4}-x^{3}-5, x_{0}=2. f^{\prime}(x)=4 x^{3}-3 x^{2}$.

Then $x_{1}=2-\frac{3}{20}=\frac{37}{20}=1.85$ and $\quad x_{2}=1.85-\frac{1.85^{4}-1.85^{3}-5}{4(1.85)^{3}-3(1.85)^{2}} \approx 1.824641$.

11. $f(x)=x-\cos (x), f^{\prime}(x)=1+\sin (x), x_{0}=0.7.$ Then $x_{1}=0.7394364978, x_{2}=0.7390851605$, and root $\approx 0.74$

13. $\frac{x}{x+3}=x^{2}-2$ so we can use $f(x)=x^{2}-2-\frac{x}{x+3}$. If $x_{0}=-4$, then the iterates $x_{n} \rightarrow-3.3615$. If $x_{0}$ $=-2$, then $x_{n} \rightarrow-1.1674$. If $x_{0}=2$, then the iterates $x_{n} \rightarrow 1.5289$.

15. $\mathrm{x}^{5}-3=0$ and $\mathrm{x}_{\mathrm{O}}=1.$ Then $\mathrm{x}_{\mathrm{n}} \rightarrow 1.2457$.

17. $f(x)=x^{3}-A$ so $f^{\prime}(x)=3 x^{2}$.

Then $x_{n+1}=x_{n}-\frac{\left(x_{n}\right)^{3}-A}{3\left(x_{n}\right)^{2}}=x_{n}-\frac{x_{n}}{3}+\frac{A}{3\left(x_{n}\right)^{2}}=\frac{1}{3}\left\{2 x_{n}+\frac{A}{\left(x_{n}\right)^{2}}\right\}$.

19. (a) $2(0)-\operatorname{INT}(2(0))=0-0=0$.

 (b) $2(1 / 2)-\operatorname{INT}(2(1 / 2))=1-1=0 .$

      $2(1 / 4)-\operatorname{INT}(2(1 / 4))=1 / 2-0=1 / 2 \rightarrow 0$

      $2(1 / 8)-\operatorname{INT}(2(1 / 8))=1 / 4-0=1 / 4 \rightarrow 1 / 2 \rightarrow 0$

       $2\left(1 / 2^{\mathrm{n}}\right)-\operatorname{INT}\left(2\left(1 / 2^{\mathrm{n}}\right)\right)=1 / 2^{\mathrm{n}-1}-0=1 / 2^{\mathrm{n}-1} \rightarrow 1 / 2^{\mathrm{n}-2} \rightarrow \ldots \rightarrow 1 / 4 \rightarrow 1 / 2 \rightarrow 0 .$

21. (a) If $0 \leq x \leq 1 / 2$, then $f$ stretches $x$ to twice its value, $2 x$.