How do you find a one-decimal place approximation for $- \sqrt[3]{35}$ $-root3 35$ ?

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How do you find a one-decimal place approximation for $- \sqrt[3]{35}$ $-root3 35$ ?

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Answer 1 · 2015-10-12T11:43:16

Consider ${(3 + t)}^{3}$ $(3+t)^3$ to find $- \sqrt[3]{35} \approx - 3.3$ $-root(3)(35)~~-3.3$

Explanation:

We know that $3^{3} = 27$ $3^3=27$ , so consider ${(3 + t)}^{3}$ $(3+t)^3$ ...

${(3 + t)}^{3} = 3^{3} + 3 \cdot 3^{2} t + 3 \cdot 3 t^{2} + t^{3}$ $(3+t)^3=3^3+3*3^2t+3*3t^2+t^3$

$= 27 + 27 t + 9 t^{2} + t^{3}$ $=27+27t+9t^2+t^3$

$\approx 27 + 27 t$ $~~27+27t$ for small values of $t$ $t$

So approximating...

$35 = 27 + 27 t$ $35=27+27t$

Hence

$t = \frac{35 - 27}{27} = \frac{8}{27} \approx 0.3$ $t=(35-27)/27=8/27~~0.3$

So $\sqrt[3]{35} \approx 3 + 0.3 = 3.3$ $root(3)(35)~~3+0.3=3.3$ and $- \sqrt[3]{35} \approx - 3.3$ $-root(3)(35)~~-3.3$

More generally, to find the cube root of a number $n$ $n$ , starting with an approximation $a_{0}$ $a_0$ , iterate using the formula:

$a_{i + 1} = a_{i} + \frac{n - a_{i}^{3}}{3 a_{i}^{2}}$ $a_(i+1) = a_i + (n - a_i^3)/(3a_i^2)$

This is a form of Newton Raphson method. Given a function $f (x)$ $f(x)$ for which we want to solve $f (x) = 0$ $f(x) = 0$ , start with an approximation $a_{0}$ $a_0$ and iterate using:

$a_(i+1) = a_i - f(a_i)/(f'(a_i))$

In our case $f(x) = x^3 - n$ and $f'(x) = 3x^2$

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How do you find a one-decimal place approximation for $- \sqrt[3]{35}$ $-root3 35$ ?

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