# Use a linear approximation to estimate the number 8.07 <mrow class="MJX-TeXAtom-ORD">

Use a linear approximation to estimate the number ${8.07}^{2/3}$
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Keely Fernandez
If you use $f\left(x\right)={x}^{2/3}$, you have $f\left(x\right)\approx f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)$. ${f}^{\prime }\left(x\right)=\frac{2}{3}{x}^{-1/3}$. If you plug in $a=8$, ${f}^{\prime }\left(8\right)=\frac{1}{3}$, so $f\left(8.07\right)=4+0.07/3=4.02333$. The real answer is $4.023299$
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Dayanara Terry
Take $f\left(x\right)={x}^{2/3}$ and $a=8$. Then
$f\left(x\right)={x}^{2/3}⇒{f}^{\prime }\left(x\right)=\frac{2}{3}{x}^{-1/3}⇒{f}^{\prime }\left(a\right)=\frac{2}{3}\cdot {8}^{-1/3}=\frac{1}{3}$
So $f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)=f\left(8\right)+\frac{1}{3}\left(.07\right)=4+\frac{1}{3}\left(.07\right)\approx 4.023$