Suppose f (x) is a differentiable function defined on the interval [0, 60]. use the function values given in the table below to answer the following:
x: 0 15 30 45 60
f(x): -3 -1 2 6 2
a) approximate the value of f '(20).
b) show that f '(c1) = 1/12 for some value x = c1 in the interval (0, 60).
c) show that f (c2) = 0 for some value x = c2 in the interval (0, 60).

For the first part: F'(20) can be approximated by finding the secant line on boundaries that hold x=20 while being as small as given. [15, 30] satisfies this. Thus F'(20) is approximated by F(30)−F(15)/30−15=−1−(−3)/15=2/15

Now for the second part: We prove b) using the mean value theorem for derivatives. SO:
x= 2 and x= -3 is -1/12. It is also given that f(x) is differentiable in [-3,2]. Wecan see from here that it is continuous at a point. Thus, we can assume that it is differentiable in [-3,2] and continuous on [-3, 2].

Now for the third part: Lastly, the intermediate value theorem allows us to prove c. Then there is a point c2 such that f(c2) = 0. We proved in the previous part that there exists a finite closed interval [a,b]. Also, there's f(30) = -1 and f(45) which satisfy the other requirement for this theorem
I hope this is what you need. 

The total would be 40.53