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

Question

anonymous · Answer

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

anonymous · Answer

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 
$$f'(27)=\frac{1}{3\sqrt[3]{27^2}}=\frac{1}{3\times 3^2}=\frac{1}{27}$$

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

alright, thx a bunch