# Thread: find cube root 0.9

1. ## find cube root 0.9

Hi again guys,

Two posts from me. But they aren't too hard. For you at least!

Given that $y = x^{-\frac{1}{3}}$ use the calculus to determine an approximate value for $\frac{1}{\sqrt [3]{0.9}}$ Normally with these there is a descernable x and $\delta x$ but I can't see a way to giggle 0.9 into anything useful. Can you?

2. I would find the first few terms of the Taylor series of $x^{-1/3}$ centered at a=1.

3. Originally Posted by s_ingram
Hi again guys,

Two posts from me. But they aren't too hard. For you at least!

Given that $y = x^{-\frac{1}{3}}$ use the calculus to determine an approximate value for $\frac{1}{\sqrt [3]{0.9}}$ Normally with these there is a descernable x and $\delta x$ but I can't see a way to giggle 0.9 into anything useful. Can you?
Use the linear approximation: $f(x + \delta x) \approx f(x) + \delta f'(x)$.

Take $f(x) = x^{-1/3}$, $x = 1$ and $\delta = -0.1$.

4. Hello, s_ingram!

Given that: $y \:=\: x^{-\frac{1}{3}}$ use the calculus to determine an approximate value for $\frac{1}{\sqrt [3]{0.9}}$
We're expected to use: . $x = 1\:\text{ and }\:dx = \text{-}0.1$

We have: . $dy \:=\:-\dfrac{1}{3}x^{-\frac{4}{3}}dx \:=\:-\frac{dx}{3\sqrt[3]{x^4}}$

If $x = 1,\:dx = \text{-}0.1$. we have: . $y + dy \:=\:1^{-\frac{1}{3}} - \frac{\text{-}0.1}{3\sqrt[3]{1^4}} \:=\:1 + \frac{1}{30} \:=\:1.0333\hdots$

Therefore: . $\frac{1}{\sqrt[3]{0.9}} \;\approx\;1.033$

5. AH!!!! So easy! I was sure that being a cube root I had to pull a 2 out, so I was playing with 2 x 9/80 = 2. 0.1125 and I wasn't getting very far! Thanks guys - you are both Fantastic!

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# how to find cuberut of 0.9

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