What is the linear approximation of the square root of 8.9? Must be accurate to the fifth decimal place.
I have tried everything it seems like yet cannot get the right answer.
2 answers
http://www.google.com/#q=square+root+of+8.9
If we let L(x) represent the linear approximation for some value of x
L(a) = f(a) + f'(a)/(x-a) , were a is a number for which I know the value of f(a)
so for √8.9
I know that √9 = 3 , so let a = 9
f(9) = √9 = 3
for f(x) = √x = x^(1/2)
f'(x) - (1/2x^(-1/2) = 1/(2√x)
so √8.9) = 3 + (1/6)(8.9-9)
= 3 - 1/60
I know 1/6 = .16666...
so 1/60 = .016666....
and √8.9 = appr 3 - .01666...
= appr 2.98333...
check with calculator: (which we are not supposed to have for this)
= 2.98329.. , not bad
L(a) = f(a) + f'(a)/(x-a) , were a is a number for which I know the value of f(a)
so for √8.9
I know that √9 = 3 , so let a = 9
f(9) = √9 = 3
for f(x) = √x = x^(1/2)
f'(x) - (1/2x^(-1/2) = 1/(2√x)
so √8.9) = 3 + (1/6)(8.9-9)
= 3 - 1/60
I know 1/6 = .16666...
so 1/60 = .016666....
and √8.9 = appr 3 - .01666...
= appr 2.98333...
check with calculator: (which we are not supposed to have for this)
= 2.98329.. , not bad
4 answers