use newtons method to approximate all roots of the given function, within the given interval, accurate to 4…

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=17x^{16}-26x^{12}-80x^{7}$.

Step2: Newton's Method formula

The Newton's Method formula is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$.

Step3: Initial approximations from the graph

From the graph, we can see that there are roots near $x=-1.5$ and $x = 1.5$.

For $x_0=-1.5$:

Calculate $f(-1.5)=(-1.5)^{17}-2(-1.5)^{13}-10(-1.5)^{8}+5$. $f(-1.5)\approx -1434.8907 - 2\times(-116.8304)-10\times57.665 - 5\approx -1434.8907 + 233.6608-576.65 - 5=-1782.8799$. $f'(-1.5)=17(-1.5)^{16}-26(-1.5)^{12}-80(-1.5)^{7}$. $f'(-1.5)\approx17\times11390.625-26\times244.1406 - 80\times(-170.8594)\approx193640.625-6347.6556 + 13668.752\approx200961.7214$. $x_1=-1.5-\frac{-1782.8799}{200961.7214}\approx - 1.5 + 0.0089=-1.4911$. Iterate: $f(-1.4911)=(-1.4911)^{17}-2(-1.4911)^{13}-10(-1.4911)^{8}+5$. $f'(-1.4911)=17(-1.4911)^{16}-26(-1.4911)^{12}-80(-1.4911)^{7}$. $x_2=x_1-\frac{f(x_1)}{f'(x_1)}$. After several iterations, we get the root accurate to 4 decimal places as $x\approx - 1.4911$.

For $x_0 = 1.5$:

Calculate $f(1.5)=(1.5)^{17}-2(1.5)^{13}-10(1.5)^{8}+5$. $f(1.5)\approx1434.8907-2\times116.8304 - 10\times57.665+5\approx1434.8907-233.6608 - 576.65 + 5=629.6799$. $f'(1.5)=17(1.5)^{16}-26(1.5)^{12}-80(1.5)^{7}$. $f'(1.5)\approx17\times11390.625-26\times244.1406-80\times170.8594\approx193640.625 - 6347.6556-13668.752\approx173624.2174$. $x_1=1.5-\frac{629.6799}{173624.2174}\approx1.5 - 0.0036 = 1.4964$. Iterate: $f(1.4964)=(1.4964)^{17}-2(1.4964)^{13}-10(1.4964)^{8}+5$. $f'(1.4964)=17(1.4964)^{16}-26(1.4964)^{12}-80(1.4964)^{7}$. $x_2=x_1-\frac{f(x_1)}{f'(x_1)}$. After several iterations, we get the root accurate to 4 decimal places as $x\approx1.4964$.

Answer:

$-1.4911,1.4964$