1a)Use Newton's Method with the function f(x)=x^2−3 and initial value x0=3 to calculate x1,x2,x3.

x1=
Be sure to use at least 6 decimal
places.

x2=
Be sure to use at least 6 decimal places.

x3=
Be sure to use at least 6 decimal places.

1b)Use Newton's Method with the function f(x)=x^3+3x−1 and initial value x0=−1.5 to calculate x1,x2,x3.

x1=
Be sure to use at least 6
decimal places.

x2=
Be sure to use at least 6 decimal
places.

x3=
Be sure to use at least 6 decimal
places.

Community Answer

Answera) Given thet f(x)=x^(2)-3,x_(0)=3 we know that new tonn methed{:[x_(n+1)=x_(n)-(f(x_(n)))/(f^(')(x_(n)))],[{:[f^(')(x)=(d)/(dx)x^(2)-(d)/(dx)3=2x],[x_(1)=x_(0+1)=x_(0)-(f(x_(0)))/(f^(')(x_(0)))]:}],[=3-(f(3))/(f^(')(3))],[f(3)=3^(2)-3=9-3=6],[f^(')(3)=2(3)=6],[x_(1)=3-(6)/(6)=3-1=2],[x_(1)=2.000000" (v) "2],[x_(2)=x_(1+1)=x_(1)-(f(x_(1)))/(f^(')(x_(1)))],[f(x_(1))=f(2)=2^(2)-3=4-3=1],[f^(')(x_(1))=f^(')(2)=2(2)=4],[x_(2)=2-(1)/(4)=(8-1)/(4)=(7)/(4)=1.75],[x_(2)=1.750000(.r)1.75]:}{:[x_(3)=x_(2)+1=x_(2)-(f(x_(2)))/(f^(')(x_(2)))],[f(x_(2))=f((7)/(4))=((7)/(4))^(2)-3=0.0625],[f^(')((7)/(4))=2((7)/(4))=(14)/(4)=(7)/(2)=3.5],[x_(3)=1.75-(0.0625)/(3.5)=1.73214285],[x_(3)=1.732143]:}b) Given that f(x)=x^(3)+3x-1,x_(0)=-1.5 (or) (-3)/(2){:[f^(')(x)=(d)/(dx)x^(3)+3(d)/(dx)x-(d)/(dx)1],[=3x^(2)+3(1)-0],[f^(')(x)=3x^ ... See the full answer