Example of a Mean Value Theorem Problem
Find the value(s) of c prescribed by the Mean Value Theorem for y= (1/3)x3 -2x2 +5x -7 over [3;6].
Solution: Since the function y is continuous over the closed interval [3;6] and dy/dx exists over the open interval (3;6), y satisfies the hypotheses of the Mean Value Theorem.
(i.e., CONNECTECD over the closed interval & SMOOTH over the open interval). Note: y = f(x)
Therefore, we may conclude the following (conclusion of the MVThm):
There exists a value c in (3;6), such that f '(c) = [f(6) – f(3)]/(6-3) (equation 1)
Let us find the value(s) of c in this problem.
Code in y or f(x) into your calculator using the menu y=
Go to 2nd graph (table) to find the values of f(6) = 23 and f(3) = -1.
We need f '(x) (which is also dy/dx): f '(x) = x2 -4x +5, so f '(c) = c2 -4c+5
Substitute into (equation 1) above to get the equation:
c2 -4c+5 = (23 – (-1))/3
or, equivalently, c2 -4c +5 =8 which gives c2-4c -3 = 0
By the quadratic formla, we get, c = [-(-4)+√(28)]/2 and c = [-(-4)-√(28)]/2
Since the 2nd value of c = [-(-4)-√(28)]/2 is not in (3;6), we reject it.
So, c = [-(-4)+√(28)]/2 = 2+√(7) (exact answer).
(approximately 4.646 to 3 decimal places)