Question

Date :
Evaluate √12 to 3 desimial places condiser the enitial guess as 4

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Answer 1

Answer:

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Answer 2

Answer:

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)=−3+6x−18

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)=−3+6x−18=6x−21

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)=−3+6x−18=6x−216x−21=0

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)=−3+6x−18=6x−216x−21=0x=216=72=3.5

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)=−3+6x−18=6x−216x−21=0x=216=72=3.5Now we iterate with x0 now being equal to 72

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)=−3+6x−18=6x−216x−21=0x=216=72=3.5Now we iterate with x0 now being equal to 722nd iteration:

Newton's method says that f(x)≈l(x)=f(x0)+f'(x0)(x−x0)√12 can be found by the function f(x)=x2−12We know that 3<√12<4, so x0=3f'(x)=2x1st iteration:l(x)=(32−12)+2(3)(x−3)=−3+6(x−3)=−3+6x−18=6x−216x−21=0x=216=72=3.5Now we iterate with x0 now being equal to 722nd iteration:l(x)=((72)2−12)

Step-by-step explanation:

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