Use a linear approximation to estimate the given number $64.07^{2/3}$

Question

That's all they give you. I tried putting it into the linear approximation equation of:

$$ f(a)+f'(a)(x-a) $$

but I get almost the same value as $64.07^{2/3}$, which is around $16.0116$. Just not sure how else to go about doing this problem. Thanks.

Here's the problem. I'm doing this online and it comes back as incorrect. I've tried nearly every possible solution to this (16.0117, 16.0116, 16.012, 16, 16.1, etc.) Is there a specific solution to this? Thanks.

Answer 1

Hint. Just rewrite $$ 64.07^{2/3}=(64+0.07)^{2/3}=64^{2/3}\left(1+\frac{0.07}{64}\right)^{2/3}=16\times\left(1+\frac{0.07}{64}\right)^{2/3} $$ and use the fact that, as $x\to 0$, $$ (1+x)^{\alpha}\sim1+\alpha x. $$

Answer 2

hint:Let $f(t) = t^{2/3}, a = 64, x = 64.07 \Rightarrow f(x) \approx f(a) + f'(a)(x-a)$. Can you continue?

Answer 3

by using Taylor series of $y=(64+x)^{2/3}$ at $x=0$ $$y=16+\frac{x}{6}-\frac{x^2}{2304}+...$$

take the firs two terms $$y=16+x/6=16+0.07/6=16.0116666$$