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this question has been giving me a little trouble:

Use a linear approximation to estimate the number $8.07^{2/3}$

I tried using $f(a)+f'(a)(x-a)$ but the answer I get ($4.02$) is apparently wrong. Any help would be appreciated!

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  • $\begingroup$ What did you use for $f,x,$ and $a$? $\endgroup$ – pwerth Dec 23 '18 at 3:21
  • $\begingroup$ Maybe you need one more digit after the decimal point $\endgroup$ – Andrei Dec 23 '18 at 3:21
  • $\begingroup$ @pwerth I used f: x^(2/3), a:8.07 $\endgroup$ – user590789 Dec 23 '18 at 3:27
  • $\begingroup$ @Andrei that didn't work either $\endgroup$ – user590789 Dec 23 '18 at 3:27
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    $\begingroup$ $a=8$, then $8^{2/3}=4$ $\endgroup$ – Andrei Dec 23 '18 at 3:38
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If you use $f(x)=x^{2/3}$, you have $f(x)\approx f(a)+f'(a)(x-a)$. $f'(x)=\frac 23 x^{-1/3}$. If you plug in $a=8$, $f'(8)=\frac 13$, so $f(8.07)=4+0.07/3=4.02333$. The real answer is $4.023299$

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Take $f(x)=x^{2/3}$ and $a=8$. Then $$f(x)=x^{2/3} \Rightarrow f'(x)=\frac{2}{3}x^{-1/3} \Rightarrow f'(a)=\frac{2}{3}\cdot 8^{-1/3}=\frac{1}{3}$$

So $f(a)+f'(a)(x-a)=f(8)+\frac{1}{3}(.07)=4+\frac{1}{3}(.07)\approx4.023$

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  • $\begingroup$ should be $x^{-1/3}$ $\endgroup$ – Andrei Dec 23 '18 at 3:44
  • $\begingroup$ @Andrei Yep, thanks $\endgroup$ – pwerth Dec 23 '18 at 3:45

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