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
$${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x – a} \right),}$$
where
$${f\left( a \right) = f\left( {27} \right) }={ \sqrt[3]{{{{27}^2}}} }={ 9.}$$
Then, the derivative using the power rule:
$${f^\prime\left( x \right) = \left( {\sqrt[3]{{{x^2}}}} \right)^\prime }={ \left( {{x^{\frac{2}{3}}}} \right)^\prime }={ \frac{2}{3}{x^{\frac{2}{3} – 1}} }={ \frac{2}{3}{x^{ – \frac{1}{3}}} }={ \frac{2}{{3\sqrt[3]{x}}}.}$$
then
$${f^\prime\left( a \right) = f^\prime\left( {27} \right) }={ \frac{2}{{3\sqrt[3]{{27}}}} }={ \frac{2}{9}.}$$
Substitute this in the equation for $L(x)$:
$${L\left( x \right) = 9 + \frac{2}{9}\left( {x – 27} \right) }={ 9 + \frac{2}{9}x – 6 }={ \frac{2}{9}x + 3.}$$
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 $f\left( x \right) = \sqrt[\large 3\normalsize]{x}$
$${f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} } = {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}$$
and its value at point $a$ is equal to
$${f’\left( {a} \right) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} } = {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}$$
Thus, getting the solution
$${f\left( x \right) \approx f\left( {a} \right) + f’\left( {a} \right)\Delta x,\;\;}\Rightarrow {\sqrt[\large 3\normalsize]{{30}} \approx \sqrt[\large 3\normalsize]{{27}} + \frac{1}{{27}} \cdot 3 } = {3 + \frac{1}{9} } = {\frac{{28}}{9} \approx 3,111.}$$
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 $\sqrt[3]{30}$?
 A: 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.
A: 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
$${f'(x) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} } = {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}$$
and its value at point $a$ equals
$${f'(a) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} } = {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}$$
Therefore since $f(a)=\sqrt[3]{27}=3$,
$$f(x)\approx L\left( x \right) ={ f\left( a \right) + f^\prime\left( a \right)\left( {x – a} \right)}=3+\frac{1}{27}\left(x-27\right).$$
Hence, the estimate for $\sqrt[3]{30}$ is
$$f(30)\approx3+\frac{1}{27}\left(30-27\right)=3+\frac{1}{9}=3.\overline{1}.$$
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.
A: 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 !
