use newtons method to approximate to three decimal places the root of the function in the given interval. x²…

Explanation:

Step1: Define the function and its derivative

Let $f(x)=x^{2}-1$, then $f'(x) = 2x$.

Step2: Choose an initial guess

We can choose an initial guess $x_0$ in the interval $[\frac{1}{2},2]$. Let $x_0 = 1.5$.

Step3: Apply Newton's method formula

Newton's method formula is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$. For $n = 0$, $x_1=x_0-\frac{f(x_0)}{f'(x_0)}$. Substitute $x_0 = 1.5$, $f(x_0)=(1.5)^{2}-1=2.25 - 1=1.25$ and $f'(x_0)=2\times1.5 = 3$. So $x_1=1.5-\frac{1.25}{3}=1.5 - 0.4167\approx1.083$. For $n = 1$, $x_2=x_1-\frac{f(x_1)}{f'(x_1)}$. $f(x_1)=(1.083)^{2}-1=1.173 - 1 = 0.173$, $f'(x_1)=2\times1.083 = 2.166$. $x_2=1.083-\frac{0.173}{2.166}=1.083 - 0.080 = 1.003$. For $n = 2$, $x_3=x_2-\frac{f(x_2)}{f'(x_2)}$. $f(x_2)=(1.003)^{2}-1=1.006 - 1=0.006$, $f'(x_2)=2\times1.003 = 2.006$. $x_3=1.003-\frac{0.006}{2.006}=1.003 - 0.003=1.000$.

Answer:

C. $x = 1.000$