The MVT states that
If f is continuous on [a,b] and differentiable on (a,b), then there exists a number c on (a,b) such that
f'(c) = [f(b) - f(a)] / b - a
In other words, the instantaneous slope must equal the average slope at some point c. (Your slope from two points will be the same as the slope of the curve at some point.)
The method I use for applying the MVT to find c on an interval is this...
(1) differentiate the function
(2) note differentiability, continuity
(3) find the regular slope (m) of the secant line
(4) set the derivative equal to m
(5) solve for c
For example, "use the MVT to find c on (-2,2) for f(x)=x^3."
(1) f'(x) = 3x^2
(2) Because f(x) is a polynomial, it is differentiable, and therefore, continuous.
(3) m = [f(2) - f(-2)] / 2 - (-2) = (8 + 8) / (2 + 2) = 16 / 4 = 4
(4) f'(c) = 3c^2 from step 1
setting it equal to our slope m=4,
f'(c) = 3c^2 = 4
(5) c^2 = 4/3
c = +-√(4/3) = +- 2/√3
I hope that helps.