Approximate $\sqrt[3]{27.3}$ by using linearization? I'm signing up for Calculus in the summer. Reading up ahead so I can understand it and get a head start. How would one go about solving this? Any hints would be appreciated.
 A: The best linear approximation to a differentiable function $f(x)$ near $x = a$ is given by $f(x) \approx f(a) + (x-a)f'(a)$. 
Now, set $f(x) = \sqrt[3]{x}$. Then, $f(27.3) \approx f(27)+0.3f'(27)$. 
Do you see where to go from here?
A: Look at the tangent line to the graph of $f(x) = \sqrt[3]{x}$ at $x = 27$. By analytic geometry, the line is $$y - f(27) = f'(27)(x - 27)$$
The tangent line is the best linear approximation near the point in question. I'm assuming you know how to differentiate $\sqrt[3]{x}$. Now, just plug in the numbers to get the line, and make $x = 27.3$ to get your approximation. 
A: Building on the answers above we can generalize this to a Taylor series
$$f(x) = \sum\limits_{i=0}^\infty \left(\frac{f^{(i)} (a)}{i!} (x-a)^i \right)$$
This can be truncated to
$$f(x) \approx \sum\limits_{i=0}^{n} \left(\frac{f^{(i)} (a)}{i!} (x-a)^i \right)$$
if $n = 1$ and $f(x) = x^{1/3}$
$$f(x) \approx a^{1/3} + \frac{x - a}{4 a^{2/3}}$$
if $a = 27$,
$$f(27.3) \approx 27^{1/3} + \frac{27.3 - 27}{4 \times 27^{2/3}}$$
$$f(27.3) \approx 3 + \frac{0.3}{4 \times 9}$$
$$f(27.3) \approx 3 + \frac{1}{120}$$
$$f(27.3) \approx \frac{361}{120}$$
if we look at $n = 2$,
$$f(x) \approx a^{1/3} + \frac{x - a}{4 a^{2/3}} + \frac{(x - a)^2}{9 a^{5/3}}$$
This improves the approximation to $f(27.3) \approx 3.01107$, which has a relative error of $8.37\times 10^{-8}$.
