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))

First, find (f'(x)) for (f(x)=x^{17}-2x^{13}-10x^{8}+5). Using the power - rule ((x^n)' = nx^{n - 1}), we have (f'(x)=17x^{16}-26x^{12}-80x^{7}).

Step2: Newton's Method formula

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 the root near (x=-1.5):

Let (x_0=-1.5). [ \begin{align*} f(x_0)&=(-1.5)^{17}-2(-1.5)^{13}-10(-1.5)^{8}+5\ f'(x_0)&=17(-1.5)^{16}-26(-1.5)^{12}-80(-1.5)^{7} \end{align*} ] [x_1=x_0-\frac{f(x_0)}{f'(x_0)}] Repeat the process (x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}) until the desired accuracy (4 decimal places) is achieved.

For the root near (x = 1.5):

Let (x_0 = 1.5). [ \begin{align*} f(x_0)&=(1.5)^{17}-2(1.5)^{13}-10(1.5)^{8}+5\ f'(x_0)&=17(1.5)^{16}-26(1.5)^{12}-80(1.5)^{7} \end{align*} ] [x_1=x_0-\frac{f(x_0)}{f'(x_0)}] Repeat the process (x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}) until the desired accuracy (4 decimal places) is achieved.

After performing the iterative calculations:

Answer:

(-1.5492,1.5492)