Question

find the linearization L(x) of the function f(x)=x^(1/3) nearby the point x=-8

Answer 1

If you calculate f'(x), you get: \[\frac{ 1 }{ 3 }x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^2}}\]
So \(f'(-8)= \dfrac{1}{3\sqrt[3]{(-8)^2}}=\dfrac{1}{3\sqrt[3]{64}} =\dfrac{1}{12}\).

This means your linear function there is \(L(x)=\dfrac{1}{12}x+b\).
If you now use the point on the graph of f with x =-8, that is: (-8, -2) and put it into the equation, you get \(b=-2 -\dfrac{1}{12} \cdot -8= ~...\)