Use linearisation of a certain function to approximate $sqrt[3]{30}$

Background

Find the linearisation of the function

$$f(x)=sqrt[3]{{{x^2}}}$$

at

$$a = 27.$$

Then, use the linearisation to find

$$sqrt[3]{30}$$

My work so far

Applying the formula

$${fleft( x right) approx Lleft( x right) }={ fleft( a right) + f^primeleft( a right)left( {x – a} right),}$$

where

$${fleft( a right) = fleft( {27} right) }={ sqrt[3]{{{{27}^2}}} }={ 9.}$$

Then, the derivative using the power rule:

$${f^primeleft( x right) = left( {sqrt[3]{{{x^2}}}} right)^prime }={ left( {{x^{frac{2}{3}}}} right)^prime }={ frac{2}{3}{x^{frac{2}{3} – 1}} }={ frac{2}{3}{x^{ – frac{1}{3}}} }={ frac{2}{{3sqrt[3]{x}}}.}$$

then

$${f^primeleft( a right) = f^primeleft( {27} right) }={ frac{2}{{3sqrt[3]{{27}}}} }={ frac{2}{9}.}$$

Substitute this in the equation for $$L(x)$$:

$${Lleft( x right) = 9 + frac{2}{9}left( {x – 27} right) }={ 9 + frac{2}{9}x – 6 }={ frac{2}{9}x + 3.}$$

Then, to use this linearisation to find

$$sqrt[3]{30}$$

I perform the following $$Delta x = x – a = 30 – 27 = 3$$ as the condition is $$x =30$$ and the staring point is $$a=27$$

As the the derivative of this particular function is given by $$fleft( x right) = sqrt[large 3normalsize]{x}$$

$${f’left( x right) = {left( {sqrt[large 3normalsize]{x}} right)^prime } } = {{left( {{x^{largefrac{1}{3}normalsize}}} right)^prime } } = {frac{1}{3}{x^{ – largefrac{2}{3}normalsize}} } = {frac{1}{{3sqrt[large 3normalsize]{{{x^2}}}}},}$$

and its value at point $$a$$ is equal to

$${f’left( {a} right) = frac{1}{{3sqrt[large 3normalsize]{{{{27}^2}}}}} } = {frac{1}{{3 cdot {3^2}}} = frac{1}{{27}}.}$$

Thus, getting the solution

$${fleft( x right) approx fleft( {a} right) + f’left( {a} right)Delta x,;;}Rightarrow {sqrt[large 3normalsize]{{30}} approx sqrt[large 3normalsize]{{27}} + frac{1}{{27}} cdot 3 } = {3 + frac{1}{9} } = {frac{{28}}{9} approx 3,111.}$$

Is my process correct so far? Or, did I go wrong in the second part? Also, as $$a=27$$ is from the original linearisation, this would be brought into the linearisation approximation for $$sqrt[3]{30}$$?

Mathematics Asked on November 12, 2021

Just to show you something a little bit different, we can do this with the binomial theorem.

$$(a+b)^k = a^k + k a^{k-1}b + frac {k(k-1)}{2} a^{k-2}b^2 + cdots$$

You learned this in algebra / pre-calculus with integers. It actually works for all real numbers.

$$(27 + 3)^frac 13 = 27^frac 13 + frac 13 (27^{-frac 23})(3) - frac 19 (27^{-frac 53})(3^2)+cdots$$

$$3 + frac 19 - frac 1{3^5}+ cdots$$

The first 2 terms would be all that you would use for a linear approximation, but for additional precision you can extend.

Answered by Doug M on November 12, 2021

The second part of your analysis is correct. The function that you want to approximate is $$f(x)=sqrt[3]{{x}}$$ at $$a=27$$, not $$g(x)=sqrt[3]{{x^2}}$$. So

$${f'(x) = {left( {sqrt[large 3normalsize]{x}} right)^prime } } = {{left( {{x^{largefrac{1}{3}normalsize}}} right)^prime } } = {frac{1}{3}{x^{ – largefrac{2}{3}normalsize}} } = {frac{1}{{3sqrt[large 3normalsize]{{{x^2}}}}},}$$

and its value at point $$a$$ equals

$${f'(a) = frac{1}{{3sqrt[large 3normalsize]{{{{27}^2}}}}} } = {frac{1}{{3 cdot {3^2}}} = frac{1}{{27}}.}$$

Therefore since $$f(a)=sqrt[3]{27}=3$$,

$$f(x)approx Lleft( x right) ={ fleft( a right) + f^primeleft( a right)left( {x – a} right)}=3+frac{1}{27}left(x-27right).$$

Hence, the estimate for $$sqrt[3]{30}$$ is

$$f(30)approx3+frac{1}{27}left(30-27right)=3+frac{1}{9}=3.overline{1}.$$

The actual value of $$sqrt[3]{30}$$ (up to the fifth decimal place) is $$3.10723$$, so the linear approximation at $$a=27$$ does quite well.

Answered by Axion004 on November 12, 2021

There is something wrong : your definition of $$f$$ is not consistant : sometimes you use $$f(x)=sqrt[3]{x^2}$$ and other times $$f(x)=sqrt[3]{x}$$... You should only use the second one I think. Get that right and you got the good idea !

Answered by Velobos on November 12, 2021

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