Question

Using first principles (the limit definition), find the tangent slope of the function y=1/(x-3) at point x=-1

Answer 1

f(x+h) - f(x) = 1/(x+h-3) - 1/(x-3) = ((x-3)-(x+h-3))/((x-3)(x+h-3)) = (-h)/((x-3)(x+h-3))

so the difference quotient is

(f(x+h)-f(x))/h = -1/((x-3)(x+h-3))
in the limit, as h→0, that is just -1/(x-3)^2
so f'(-1) = -1/16

Or, you can start out as (f(-1+h) - f(-1))/h and follow the same steps.