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I am having problems with the following question:

Use the linear approximation $(1+x)^k\approx 1+kx$ to find an approximation for the function $f(x)$ for values of $x$ near zero $$f(x)=\sqrt[3]{\left(1-\frac1{2+x}\right)^2}.$$

When $x$ approaches $0$, the fraction part approaches $\frac12$, which is far from $0$. So I wonder how I can apply the approximation formula given by the question to evaluate the linear approximation of $f(x).$

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    $\begingroup$ I have typed out your question - please refrain from posting images in the future as not all users can see them $\endgroup$ – lioness99a Jan 25 at 13:41
  • $\begingroup$ Thank you a lot :) $\endgroup$ – Jack Hwo Jan 25 at 13:50
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$$ f(x) = \left( 1- \frac{1}{2+x}\right)^{2/3} = \left( 1- \frac{1}{2} \frac{1}{1+x/2}\right)^{2/3} = \left( 1- \frac{1}{2} (1+x/2)^{-1}\right)^{2/3} $$ and $(1+x/2)^{-1} \approx 1-x/2$ so therefore $$ f(x) \approx \left( 1- \frac{1}{2}(1-x/2)\right)^{2/3} $$ ... and so on. Maybe like this?

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  • $\begingroup$ That is not a linear approximation ... lol $\endgroup$ – Jack Hwo Jan 25 at 13:53
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    $\begingroup$ I have presented my idea to you, one which you can continue. It's not ready yet. $\endgroup$ – Matti P. Jan 25 at 13:54
  • $\begingroup$ Okay, thank you :) $\endgroup$ – Jack Hwo Jan 25 at 13:56
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Maybe like this? $$ f(x) = \left( 1 - \frac 1 {2+x}\right)^{2/3} \approx 1 - \frac 2 {3(2+x)} $$

Edit: indeed, this works when $\frac 1 {2+x}$ is small, which isn't the case when $x$ is close to zero.

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    $\begingroup$ In this problem, we are looking at $x \approx 0$, not $\frac{1}{2+x}\approx 0$. I also almost fell into this trap ... $\endgroup$ – Matti P. Jan 25 at 13:58
  • $\begingroup$ Yep, that's the answer the solution offers. But I wonder why... $\endgroup$ – Jack Hwo Jan 25 at 14:01
  • $\begingroup$ The approximation $(1 + y)^k \approx 1+yk$ holds when $y$ is near zero; you are applying it in a case where $y = -\frac1{2+x}$ is near $-1/2$. $\endgroup$ – Mees de Vries Jan 25 at 14:03
  • $\begingroup$ Thank you, I am now able to understand the problem. I think the question doesn't make sense a lot. Next time, I'd rather use derivatives to figure out the approximation. Any way, thank you:) $\endgroup$ – Jack Hwo Jan 25 at 14:14
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The simplest linear model could come from composition of Taylor series built at $x0$. You have $$1-\frac{1}{x+2}=\frac{1}{2}+\frac{x}{4}-\frac{x^2}{8}+O\left(x^3\right)$$ Now, still using Taylor or binomial expansion $$\left(1-\frac{1}{x+2}\right)^{2/3}=\frac{1}{2^{2/3}}+\frac{x}{3\ 2^{2/3}}-\frac{7 x^2}{36\ 2^{2/3}}+O\left(x^3\right)=\frac{1}{2^{2/3}}+\frac{x}{3\ 2^{2/3}}+O\left(x^2\right)$$ which is not bad for $0.0 \leq x \leq 0.2$.

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