Question

The graph of function f contains the point P(1, 2) and Q(s, r). The equation of the secant line through P and Q is y = ((s² + 2s - 3)/(s - 1))x - 1 - s. The value of f' (1), is

(A) 2 

(B) 3 

(C) 4 

(D) non existent  

Answer 1

Answer is (C) 4

I By definition f'(1) is the limit of the slope of the secant line when s →1.

By substituting x = s into the equation of the secant line, and cancelling by s - 1 again, we get y = s² + 2s -1. 

This is f(s), and its derivative is f'(s) = 2s + 2' so f'(1) = 4.