## University Calculus: Early Transcendentals (3rd Edition)

$L(x)=\frac{x}{12}+\frac{4}{3}$
$f(x)=(x)^{\frac{1}{3}}$, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=\frac{1}{3}{(x)^{\frac{-2}{3}}}$ The value of$x_{0}$ is chosen close to 8.5; we choose the value of 8 for simplicity $x_{0}=8$ $f(8)=(8)^{\frac{1}{3}}=2$ $f'(8)=\frac{1}{3}{(8)^{\frac{-2}{3}}}=\frac{1}{12}$ then $L(x)=2+\frac{1}{12}(x-8)=\frac{x}{12}+\frac{4}{3}$ $L(x)=\frac{x}{12}+\frac{4}{3}$ the final answer :$L(x)=\frac{x}{12}+\frac{4}{3}$