Question

what is the equation of the tangent line to the graph of f(x)= x^3+1, at the point (1,2)?

Answer 1

(y-2)={d/dx (x^3 +1) at (1,2)} (x-1)
=> y-2 = 3(x-1)

Answer 2

f'(x) = 3x^2 ==> slope = 3 ===> y=3x-1

Answer 3

can you use the limit process?

Answer 4

yes you can....
f '(x=)lim(h->0)={[(x+h)^3 +1]-[(x)^3+1]}/h
f '(x) = 3x^2 slope=3
y-2 = 3(x-1)
y=3x-1