Answer
$L(x)=\frac{1}{12}x-\frac{4}{3}$
Work Step by Step
Step 1. $f(x)=\sqrt[3] x=(x)^{1/3}$, $f'(x)=\frac{1}{3}x^{-2/3}=\frac{1}{3\sqrt[3] {x^2}}$,
Step 2. $f(-8)=-2$, $f'(-8)=\frac{1}{3\sqrt[3] {(-8)^2}}=\frac{1}{12}$,
Step 3. The linearization at $x=-8$ is $L(x)=f(-8)+f'(-8)(x+8)=-2+\frac{1}{12}(x+8)=\frac{1}{12}x-\frac{4}{3}$
