
Linear Approximation
Find the linear approximation, for...
$\displaystyle y=mx+b, to f(x) = \sqrt[3]{3x1}$
at the point x=3
Ok, so
$\displaystyle f(3)=2$
$\displaystyle f'(x)=3x1$
$\displaystyle f'(3)=8$
now to solve for y=mx+b?..
$\displaystyle 2=8(3)+b$
$\displaystyle b=22$
$\displaystyle y=8x22$
this is apparently wrong. What am I doing wrong here?

I think you got the wrong expression for f'(x)
$\displaystyle f(x) = (3x  1)^{\frac13}$
$\displaystyle f'(x) = \dfrac13 (3x1)^{\frac23} \cdot 3 = (3x1)^{\frac23} = \sqrt[3]{\dfrac{1}{(3x1)^2}}$