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?

Answer 1

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.

Answer 2

Oh! Minor mistake, thanks for the help!