# How to do this Linear Approximation?

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!

• What did you use for $f,x,$ and $a$? – pwerth Dec 23 '18 at 3:21
• Maybe you need one more digit after the decimal point – Andrei Dec 23 '18 at 3:21
• @pwerth I used f: x^(2/3), a:8.07 – user590789 Dec 23 '18 at 3:27
• @Andrei that didn't work either – user590789 Dec 23 '18 at 3:27
• $a=8$, then $8^{2/3}=4$ – Andrei Dec 23 '18 at 3:38

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 $$4.023299$$

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

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