Question
Use differentials to approximate the value of the expression.
cube root of 25
I set x=27 and dx=-2 and got ~3.0741, what did I do wrong?
cube root of 25
I set x=27 and dx=-2 and got ~3.0741, what did I do wrong?
Answers
Reiny
our f(x) = x^(1/3) , and
f ' (x) = (1/3)x^(-2/3) = 1/(3(x)^(2/3) )
I will go with your x = 27 , dx = -2
f(x + dx) is appr = f(x) + f ' (x) (dx)
f(25) = f(27) + (1/3)(27)^(-2/3)
= 3 + (1/27)(-2)
= 3 - .0741
= 2.926
you simply added the part at the end instead of subtracting it. Looks like you did the rest ok
(I would give you most of the marks)
notice calculator answer is 2.924, so our approximation is pretty good.
f ' (x) = (1/3)x^(-2/3) = 1/(3(x)^(2/3) )
I will go with your x = 27 , dx = -2
f(x + dx) is appr = f(x) + f ' (x) (dx)
f(25) = f(27) + (1/3)(27)^(-2/3)
= 3 + (1/27)(-2)
= 3 - .0741
= 2.926
you simply added the part at the end instead of subtracting it. Looks like you did the rest ok
(I would give you most of the marks)
notice calculator answer is 2.924, so our approximation is pretty good.
Ray
Oh! Minor mistake, thanks for the help!