Question

Need help with linear approximation...
f(x) = x^1/3, find equation of tangent line when x = 27.

Answer 1

take the derivative
it is
\[f'(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^2}}\]

Answer 2

then for the point on the graph find
\[f(27)=\sqrt[3]{27}=3\] so the point is (27,3) and to find the slope find
\[f'(27)=\frac{1}{3\sqrt[3]{27^2}}=\frac{1}{3\times 3^2}=\frac{1}{27}\]

Answer 3

then use the point slope formula to find the equation for the line

Answer 4

can you show me that part as well? I'm kinda lost in this part as well.

Answer 5

use
\[y-y_1=m(x-x_1)\] with
\[m=\frac{1}{27}, x_1=27,y_1=3\]
a direct substitution will give you the equation for the line

Answer 6

alright, thx a bunch