use newtons method to approximate the indicated root of the equation correct to six decimal places. the…

Explanation:

Step1: Define the function and its derivative

Let (f(x)=x^{2}-5\sin x). Then (f^{\prime}(x) = 2x - 5\cos x).

Step2: Choose an initial guess

By graphing (y = 5\sin x) and (y=x^{2}) (or by inspection), we can choose an initial guess (x_{0}=2).

Step3: Apply Newton's method formula

Newton's method formula is (x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}). For (n = 0): (f(2)=2^{2}-5\sin(2)\approx4 - 5\times0.9093=-0.5465) (f^{\prime}(2)=2\times2-5\cos(2)\approx4+5\times0.4161 = 6.0805) (x_{1}=x_{0}-\frac{f(x_{0})}{f^{\prime}(x_{0})}=2-\frac{- 0.5465}{6.0805}\approx2 + 0.0899 = 2.0899) For (n = 1): (f(2.0899)=(2.0899)^{2}-5\sin(2.0899)\approx4.3677-5\times0.8979=4.3677 - 4.4895=-0.1218) (f^{\prime}(2.0899)=2\times2.0899-5\cos(2.0899)\approx4.1798 + 5\times0.4404=4.1798+2.202=6.3818) (x_{2}=2.0899-\frac{-0.1218}{6.3818}\approx2.0899 + 0.0191=2.109) For (n = 2): (f(2.109)=(2.109)^{2}-5\sin(2.109)\approx4.4478-5\times0.8894=4.4478 - 4.447=0.0008) (f^{\prime}(2.109)=2\times2.109-5\cos(2.109)\approx4.218+5\times0.4571=4.218 + 2.2855=6.5035) (x_{3}=2.109-\frac{0.0008}{6.5035}\approx2.109 - 0.0001=2.1089)

Answer:

(2.108900)