hello,
f(x ) = x / (x-2) x is [-4,-1]
f'(x ) = [f(-1) - f(-4)] / ( -1 -(-4))
f(-1) = -1/-3 = 1/3
f(-4) = -4/-6 = 2/3
f'(x) = [f(-1) - f(-4)] / ( -1 -(-4)) = [1/3 - 2/3 ] / 3 = -1/9
Is it possible to determine the instantaneous rate of change at x=2?
No, because this function at x = 2 is not continuous and x = 2 is vertical asymptote.
Given the function g(x)=x4-2x3+1
estimate the instantaneous rate of change at x=1
g'(x) = [g(x) - g(1)] / (x-1)
g(x)=x4-2x3+1
g(1) = 1-2+1 =0
g'(x) = [g(x) - g(1)] / (x-1) = [x4-2x3+1- 0 ] / (x-1) = (x-1) ( x^3 -x^2 -x -1 ) / (x-1) = x^3 -x^2 -x -1= -2
On the other hand: g'(x) = 4x^3 -6x^2 ------> g'(1 ) = 4 - 6 = -2 the same answer
I hope it is useful,
Minoo