Asked by Ray

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?

Answers

Answered by 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.
Answered by Ray
Oh! Minor mistake, thanks for the help!
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