use newtons method to approximate the real zeros of each function. continue until two successive…

Explanation:

Step1: Find the derivative of the function

The function is $y = -x^{5}+2x^{3}+2$. Using the power - rule $(x^n)^\prime=nx^{n - 1}$, the derivative $y^\prime=f^\prime(x)=-5x^{4}+6x^{2}$.

Step2: Apply Newton's Method formula

Newton's Method formula is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^\prime(x_{n})}$. Let $x_0 = 1.8$. First, calculate $f(x_0)$ and $f^\prime(x_0)$: $f(x_0)=-(1.8)^{5}+2(1.8)^{3}+2=-18.89568 + 11.664+2=-5.23168$. $f^\prime(x_0)=-5(1.8)^{4}+6(1.8)^{2}=-5\times10.4976+6\times3.24=-52.488 + 19.44=-33.048$. Then $x_1=x_0-\frac{f(x_0)}{f^\prime(x_0)}=1.8-\frac{-5.23168}{-33.048}\approx1.8 - 0.1583=1.6417$.

Step3: Iterate

Calculate $f(x_1)$ and $f^\prime(x_1)$: $f(x_1)=-(1.6417)^{5}+2(1.6417)^{3}+2$. $f(x_1)\approx-11.57+8.77+2=-0.8$. $f^\prime(x_1)=-5(1.6417)^{4}+6(1.6417)^{2}$. $f^\prime(x_1)\approx-5\times7.29+6\times2.69=-36.45+16.14=-20.31$. $x_2=x_1-\frac{f(x_1)}{f^\prime(x_1)}=1.6417-\frac{- 0.8}{-20.31}\approx1.6417 - 0.0394=1.6023$.

Step4: Continue iteration

Calculate $f(x_2)$ and $f^\prime(x_2)$: $f(x_2)=-(1.6023)^{5}+2(1.6023)^{3}+2\approx-8.29+8.29+2 = 2$. $f^\prime(x_2)=-5(1.6023)^{4}+6(1.6023)^{2}\approx-5\times4.11+6\times2.57=-20.55 + 15.42=-5.13$. $x_3=x_2-\frac{f(x_2)}{f^\prime(x_2)}=1.6023-\frac{2}{-5.13}\approx1.6023 + 0.3899=1.9922$.

Step5: Keep iterating

Calculate $f(x_3)$ and $f^\prime(x_3)$: $f(x_3)=-(1.9922)^{5}+2(1.9922)^{3}+2\approx-31.6+15.8+2=-13.8$. $f^\prime(x_3)=-5(1.9922)^{4}+6(1.9922)^{2}\approx-5\times15.7+6\times3.97=-78.5+23.82=-54.68$. $x_4=x_3-\frac{f(x_3)}{f^\prime(x_3)}=1.9922-\frac{-13.8}{-54.68}\approx1.9922-0.2524 = 1.7398$.

Step6: Another iteration

Calculate $f(x_4)$ and $f^\prime(x_4)$: $f(x_4)=-(1.7398)^{5}+2(1.7398)^{3}+2\approx-15.7+10.4+2=-3.3$. $f^\prime(x_4)=-5(1.7398)^{4}+6(1.7398)^{2}\approx-5\times8.7+6\times3.02=-43.5+18.12=-25.38$. $x_5=x_4-\frac{f(x_4)}{f^\prime(x_4)}=1.7398-\frac{-3.3}{-25.38}\approx1.7398 - 0.13=1.6098$.

Step7: One more iteration

Calculate $f(x_5)$ and $f^\prime(x_5)$: $f(x_5)=-(1.6098)^{5}+2(1.6098)^{3}+2\approx-8.6+8.4+2=1.8$. $f^\prime(x_5)=-5(1.6098)^{4}+6(1.6098)^{2}\approx-5\times4.3+6\times2.6=-21.5+15.6=-5.9$. $x_6=x_5-\frac{f(x_5)}{f^\prime(x_5)}=1.6098-\frac{1.8}{-5.9}\approx1.6098 + 0.3051=1.9149$.

Step8: Keep going

Calculate $f(x_6)$ and $f^\prime(x_6)$: $f(x_6)=-(1.9149)^{5}+2(1.9149)^{3}+2\approx-25.9+13.9+2=-10$. $f^\prime(x_6)=-5(1.9149)^{4}+6(1.9149)^{2}\approx-5\times13.0+6\times3.67=-65+22.02=-42.98$. $x_7=x_6-\frac{f(x_6)}{f^\prime(x_6)}=1.9149-\frac{-10}{-42.98}\approx1.9149-0.2327 = 1.6822$.

Step9: Final iteration

Calculate $f(x_7)$ and $f^\prime(x_7)$: $f(x_7)=-(1.6822)^{5}+2(1.6822)^{3}+2\approx-12.8+9.7+2=-1.1$. $f^\prime(x_7)=-5(1.6822)^{4}+6(1.6822)^{2}\approx-5\times6.5+6\times2.83=-32.5+16.98=-15.52$. $x_8=x_7-\frac{f(x_7)}{f^\prime(x_7)}=1.6822-\frac{-1.1}{-15.52}\approx1.6822-0.0709 = 1.6113$. $x_9=x_8-\frac{f(x_8)}{f^\prime(x_8)}$. Calculate $f(x_8)$ and $f^\prime(x_8)$: $f(x_8)=-(1.6113)^{5}+2(1.6113)^{3}+2\approx-8.7+8.5+2=1.8$. $f^\prime(x_8)=-5(1.6113)^{4}+6(1.6113)^{2}\approx-5\times4.4+6\times2.6=-22+15.6=-6.4$. $x_9=1.6113-\frac{1.8}{-6.4}\approx1.6113 + 0.2813=1.8926$. $x_{10}=x_9-\frac{f(x_9)}{f^\prime(x_9)}$. Calculate $f(x_9)$ and $f^\prime(x_9)$: $f(x_9)=-(1.8926)^{5}+2(1.8926)^{3}+2\approx-24.7+13.4+2=-9.3$. $f^\prime(x_9)=-5(1.8926)^{4}+6(1.8926)^{2}\approx-5\times12.2+6\times3.58=-61+21.48=-39.52$. $x_{10}=1.8926-\frac{-9.3}{-39.52}\approx1.8926-0.2353 = 1.6573$. $x_{11}=x_{10}-\frac{f(x_{10})}{f^\prime(x_{10})}$. Calculate $f(x_{10})$ and $f^\prime(x_{10})$: $f(x_{10})=-(1.6573)^{5}+2(1.6573)^{3}+2\approx-11.2+9.1+2=-0.1$. $f^\prime(x_{10})=-5(1.6573)^{4}+6(1.6573)^{2}\approx-5\times7.0+6\times2.75=-35+16.5=-18.5$. $x_{11}=1.6573-\frac{-0.1}{-18.5}\approx1.6573-0.0054 = 1.6519$. $x_{12}=x_{11}-\frac{f(x_{11})}{f^\prime(x_{11})}$. Calculate $f(x_{11})$ and $f^\prime(x_{11})$: $f(x_{11})=-(1.6519)^{5}+2(1.6519)^{3}+2\approx-10.9+8.9+2=0$. $f^\prime(x_{11})=-5(1.6519)^{4}+6(1.6519)^{2}\approx-5\times6.7+6\times2.73=-33.5+16.38=-17.12$. $x_{12}=1.6519-\frac{0}{-17.12}=1.6519$. Since $|x_{12}-x_{11}|=|1.6519 - 1.6519|=0<0.001$.

Answer:

$1.6519$