Mathematics Asked on November 12, 2021
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 $sqrt[3]{30}$?
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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