How to Find Value of Root √5,√7,√11
adamsyakir:
hmm wait..
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Use polinom Taylor orde 1 method
f₁(x)=f(a)+f'(a)(x-a)
the value of a is approximation of x.
Use the function f(a)=√a,so f'(a)= 1 (derivative)
2√a
⇒√5 →is x
a=4 because it is approaching ,f(a)=√4=2
f'(a)= 1 = 1 =0,25
2√4 4
So,
f₁(5)=2 +0,25(5-4)
=2+0,25
=2,25
------------------------------------------------------------------------------------------
⇒√7 →is x
a=4 because approaching,f(a)=√4=2
f'(a)= 1 = 1 =0,25
2√4 4
So,
f₁(7)=2+0,25(7-4)
=2+0,25(3)
=2,75
-------------------------------------------------------------------------------------------
⇒√11 →is x
a=9 because approaching ,f(a)=√9=3
f'(a)= 1 = 1
2√9 6
So,
f₁(11)=3+ 1.(11-9)
6
=3+1.(2)
6
=3+2
6
=3+1
3
=3+0,33
=3,33
OR
=3 x 3+1
3
=10
3
f₁(x)=f(a)+f'(a)(x-a)
the value of a is approximation of x.
Use the function f(a)=√a,so f'(a)= 1 (derivative)
2√a
⇒√5 →is x
a=4 because it is approaching ,f(a)=√4=2
f'(a)= 1 = 1 =0,25
2√4 4
So,
f₁(5)=2 +0,25(5-4)
=2+0,25
=2,25
------------------------------------------------------------------------------------------
⇒√7 →is x
a=4 because approaching,f(a)=√4=2
f'(a)= 1 = 1 =0,25
2√4 4
So,
f₁(7)=2+0,25(7-4)
=2+0,25(3)
=2,75
-------------------------------------------------------------------------------------------
⇒√11 →is x
a=9 because approaching ,f(a)=√9=3
f'(a)= 1 = 1
2√9 6
So,
f₁(11)=3+ 1.(11-9)
6
=3+1.(2)
6
=3+2
6
=3+1
3
=3+0,33
=3,33
OR
=3 x 3+1
3
=10
3
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