What is the average rate of change of f(x) given f(x)= -6/x from (1,2) and (1,4)

what is the instantaneous rate of change of f(x) x=1

I'm going to assume (1,2) represents the interval from (1,2), although I think perhaps it should be written as [1,2].

f(x)= -6/x
f(1) = -6/1 = -6
f(2) = -6/2 = -3
Avg. Rate of Change = m = change in y/change in x = -6 - (-3) / 1 - 2 = -3 / -1 = 3

f(x) = -6/x
f(1) = -6/1 = -6
f(4) = -6/4 = -3/2
Avg. Rate of change = m = change in y/change in x = -(3/2) - (-6) / 4 - 1 = (9/2) / 3 = 1.5

If you are supposed to just guess the instantaneous rate of change at x=1, it looks like as the secent between the two points gets closer and closer to the tangent line at x=1, the rate of change gets close to some amount a little bit greater than 3. It would probably be responsible to guess about 5.

If you are supposed to find the exact answer, we can use the difference quotient:
lim(h-->0) [f(x+h) - f(x)] / h
lim(h-->0) [-6/(x+h) - (-6/x)] / h
lim(h-->0) [ (-6x + 6x + 6h)/(x^2+xh) ] / h
lim(h-->0) [(6h)/(x^2+xh)] / h
lim(h-->0) 6 / (x^2 + xh)
f'(x) = 6 / x^2
f'(1) = 6 / 1^2
= 6