Mean Value Theorem

Question

Consider the function f(x)=1/x on the interval [3,12]. Find the average or mean slope of the function on this interval.  By the Mean Value Theorem, we know there exists a c in the open interval (3,12) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.

Answer 1

For f(x) = x^-1 or 1/x {with f'(x) = -x^-2 or -1/x²} continuous on closed interval [3, 12] and differentiable on open interval (3, 12), there is a value c in (3, 12) such that f'(c) (that is, -c^-2 or -1/c²) is equal to [f(12) − f(3)]/(12 − 3).

Then write -1/c² = [1/12 − 1/3]/(12 − 3) or [-3/12]/9 or -1/36. This goes to c² = 36; c in (3, 12) must be 6.