Find the slope of the tangent line to the function f(x)=1/x-3 at a=−1.

Answer 1

Express f(x) = 1/(x − 3) as (x − 3)^-1.

Then f'(x) or d(x − 3)^-1/dx is constructed as (-1)(1)(x − 3)^{-1 − 1} or -(x − 3)^-2.

Assuming that the statement of the problem above means to say "the slope of the line tangent to the graph of f(x) at x = -1", evaluate f'(-1) or -(-1 − 3)^-2 which goes to -1/(-4)² or -1/16 (equal to -0.0625).